Nuclear, Particle, and Atomic Physics

QUESTION:
I recently learned that at CERN they produced streams of protons by applying large magnetic fields to hydrogen gas. This removes the electrons and these only the nucleus specifically protons. This made me wonder as alpha particles are merely the nucleus of the helium gas, were you to apply a large magnetic field to helium gas would you merely be left with alpha particles?

ANSWER:
I don't where you "learned" this, but it is almost 100% wrong. In a proton accelerator you need to do three things:

Ionize hydrogen gas to make a plasma. This is done with a radio frequency electric field.
You then extract the protons from this plasma using electric fields and inject those into an accelerator which speeds them up.
Then you repeatedly speed them up, again using electric fields and magnetic fields to steer and focus the protons being accelerated.

Most accelerators, including for alpha particles, work the same way. In fact, magnetic fields cannot ever exert the forces necessary to accelerate charged particles. The reason is that the force which a magnetic field B exerts on a charged particle with a velocity v is perpendicular to v.

QUESTION:

do the proton and the neutron have exactly the same mass?
how do masses of the proton and neutron compared to the mass of the electron? w
which particles make the greatest contribution to the mass of an atom?
which particles make the greatest contribution to the chemical properties of an atom?

ANSWER:
I usually do not answer multiple questions, but these are pretty much one question about properties of atoms.

The neutron is about 0.14% heavier than the proton.
The electron is about 1837 times lighter than a proton.
All stable atoms but ¹H have neutrons as having to largest fraction of the weight. All stable nuclei except hydrogen have more or an equal number of neutrons than protons.
The electrons, which orbit the nucleus comprised of protons and neutrons, determine the chemical properties of the atom

QUESTION:
Recently in my university, we had to do an experiment on beta spectroscopy. They gave us a sample of radioactive strontium-90 and told us that it decays beta particles at energies of 0.546MeV and 2.274 MeV and below. After finishing the experiment and submitting the report, the lab instructors told me that i had understood the concept wrongly. I had thought that at the energies of 0.546MeV and 2.274 MeV i would find a high number of electrons emitted. They told me that this was not the case and that at these energies you would expect to find no electrons emitted. They told me that electrons were not emitted at this energies because the neutons would absorb all the energy and leave the electrons in the sample thus no electrons would be detected. However, my understanding is that at these supposed max energies shouldn't the eletrons be emitted and leave the atom behind, allowing us to detect it? Why does the electron not leave the atom when it has high energies? Also during the experiment if there is supposed to be no electrons detected why did i detect a lot of electrons at these energies that is orders of magnitudes higher than that of the background radiation?

ANSWER:
Let's first talk about what happens in β^- decay: a nucleus with too many neutrons to be stable changes a neutron to a proton (which stays in the nucleus), an electron, and a neutrino. The energy released by this decay is therefore shared between the electron and the neutrino (assuming that the nucleus, being enormously heavier than the other two particles, will have almost no recoil energy. The electron, as you demonstrate in your lab assignment, is pretty easy to observe. The neutrino, however, is extremely difficult to detect—the sun produces copious amounts of them and nearly all pass completely through the earth without interacting at all. When β^- decay was first discovered, it was a huge mystery because the electrons, the only thing observed, were not emitted with a single energy but with a broad spectrum of energies. A typical electron spectrum is shown in the graph to the left; in that graph you see a maximum near 0.1 MeV electron energy but, in this example, the total energy should be about 1.15 MeV. There seemed to be only two explanations: either energy is not conserved or else there is a third particle which was not observed and carries off some of the energy. Everyone believed the second explanation and the existence of neutrinos was totally accepted by all physicists. They were so hard to observe that it was 26 years after Wolfgang Pauli proposed them (1930) and they were observed by Cowan and Reines (1956).

In your experiment, there were two decays going on simultaneously: ⁹⁰Sr decaying to ⁹⁰Y and ⁹⁰Y decaying to ⁹⁰Zr. Both have a maximum but they are far smaller than the total energies of the decays. Nearly nothing after the first sentence starting with "They told me…" makes little or no sense but that is probably because you did not understand what they were trying to tell you.

QUESTION:
If the space inside an atom is totally empty, therefore a vacuum, why doesn't the atom collapse?

ANSWER:
If the space inside the solar system is totally empty, therefore a vacuum, why doesn't the solar system collapse? Because there is a force, gravity, pointing from the object to the center of the sun which keeps all the moving objects in the solar system moving in elliptical orbits. The same is true of the Bohr model of the atom except the force is the electrostatic force rather than gravity. You should be aware, however, that the inside of an atom is not a vacuum. The electrons are not really in simple orbits as the Bohr model would have it, but they have wave functions (distributions of the probabilities of their being at any place in space) which extend all the way into the nucleus of the atom.

QUESTION:
a light photon moving at the speed of light, can it have spin? and if so can the spin be added to its speed giving it a faster than light speed breaking the laws of physics? I have always wondered this but never seem to find out if they have spin....

ANSWER:
Every photon has a spin of 1; this means that it has an intrinsic spin angular momentum of L=√[1(1+1)]ℏ. So, I am guessing that you are visualizing the photon as a little spinning ball and think that you can conclude (see my figure) that the "equator" on the near side of the ball is moving forward with a speed of c+v. However, spin in quantum mechanics cannot be visualized using such a simple classical model, even though we often do think of spin this way to get a qualitative feel for spin. For example, if you visualize an electron as a spinning uniform sphere with a reasonable radius, you find that the surface must have a speed greater than the speed of light.

QUESTION:
I recently watched with great interest a PBS program which described in layman's terms how Uranium 238 transforms into the different chained elements, to include U235. It also explained the basics of the chain reaction caused by splitting U235 using E=MC^2 as the basis for energy release. This is where I was a little unclear. The split was described as one U325 nucleus splitting into two separate nuclei with some individual particles released (can't remember if they are protons or neutrons) Those particles then collide with other U235 atoms in proximity triggering subsequent splits and particle releases as part of a chain reaction. My question is that the mass doesn't appear to be transforming into energy (E=MC^2). Rather it appears that it is simply splitting and being cast off, so what causes the energy release? This assumes that the number of particles in the remaining two nuclei + the particles independently released still equal 235. There was a general reference in the program to how the Strong Force reduces the size of the resultant smaller nuclei, but it didn't say if matter within each was converted to energy or if the number of particles are additionally reduced through such a conversion. Thanks for any clarification you can provide.

ANSWER:
Suppose that you weigh one ²³⁵U and one neutron. Now, when you add the neutron to the ²³⁵U it fissions and, after all is said and done you have two lighter atoms and a few neutrons; if you weigh all these byproducts, you will find that approximately 0.1% of the original weight is missing. Where did it go? Most of it went into kinetic energy of the byproducts, that is they are all moving faster. Kinetic energy of atoms is essentially what thermal energy is—the reactor (or bomb) has gotten hotter. For a bomb it all gets enormously hotter resulting in the explosion. For a reactor, the rate of fissioning is controlled and the heat is extracted to drive turbines to create electricity. More detail can be found in an earlier answer.

FOLLOWUP QUESTION:
I'm assuming (correct me if I'm wrong) that such motion was some percent of the speed of light which would account for the transformation of about 0.1% of its matter to energy.

ANSWER:
First things first—there was an error in my original answer, now corrected throughout: the amount of mass converted to energy is about 0.1%; nuclear fusion is about 1%. No, the motion of the atoms is nowhere near the speed of light. It is simply classical ½mv² type of kinetic energy. The "transformation" is simply that—mass energy transformed into kinetic energy. To understand, see a recent answer which explains why a bound system has less mass than if it is pulled apart and the mass measured. It turns out that heavy nuclei like uranium are less tightly bound than nuclei with roughly half their mass; therefore when they split the products are less massive. That is why fission works as an energy source.

QUESTION:
I don't know if you know about the device scientist called the torus experiment. I heard about it at UGA thirty years ago. It was to make a round tube with wire around all of it. Then special particles went through the tube, and the device put out a lot of energy. They tried to make one 5 miles in diameter, but they kept having problems with centrifugal force. The machine would make as much electricity as a nuclear power plant, or more. What I thought of is real simple, but maybe it was overlooked or something. My idea is to make the shape like a track around a football field, and make the straight parts a lot longer. So the particles would have time to realign in the center and all start going the same speed. My teacher said it is like a contained nuclear explosion. It would put out a lot more electricity than it takes to run it.

ANSWER:
What you are describing sounds like a tokamak, an experimental fusion reactor. Nuclear fusion is the energy which powers stars. Roughly, the idea is that if you get hydrogen atoms hot enough (hundreds of millions of degrees) they will have enough energy to fuse together to make helium, and that releases a great deal of energy. The problem turned out to be much more challenging than anticipated and there still is not a practicable fusion reactor which will give you more energy out than you use to run it. The main problem was the difficulty of keeping the hot plasma confined long enough to have a sufficient number of fusion reactions. The current research is centered on constructing a tokamak by an international collaboration known as ITER; the lab is in France. There is an old cynical joke among physicists: "Fusion is the energy fo the future and always will be!"

QUESTION:
I am really confused over the spin of neutrino, my teacher says it is always spin 1/2, however with my calculation from beta decay it can be 3/2 also? Please clarify why it can't be spin 3/2?

ANSWER:
I have no idea how you can calculate the spin of a particle from the decay. Spin is the quantum number of the intrinsic angular momentum of a particle, one of the properties of the particle itself, and cannot be anything other than what it is, ½ in the case of a neutrino (or electron, or proton, or neutron). A spin-½ particle might also have an orbital angular momentum quantum number L and its total angular momentum quantum number J=L±½ could be different from its spin. Maybe that is how you are coming up with 3/2. For example, if L=1 for the neutrino, J=½ or 3/2. (By the way, if you are just learning about angular momentum, the actual angular momentum of an object with angular momentum quantum number N is (h/2π)√(N(N+1)) where h is Planck's constant.)

QUESTION:
Since the rest mass of the constituent quarks only makes up about 2% of the total mass of the proton (or neutron), would you please explain in layman's terms what it is in QCD that forces all protons (or neutrons) to have the exact same mass. Does the answer to this question imply that time is quantized?

ANSWER:
One of the things which the uncertainty principle says is that you cannot know precisely both the energy and the time of a quantum system, ΔEΔt≈h/2π where h/2π=6.6x10^-16 eV·s is the rationalized Planck's constant. In order to measure the energy of a system, you must spend some time; the more time you have to examine the system, the more accurately you can determine its mass. But, if the system is unstable, that is it has some lifetime before it decays to a different state, you will get a different answer every time you try to measure its mass (which is, equivalently, its energy). A proton is a stable particle and therefore, since it lives forever its mass has zero uncertainty and is why all protons have identically the same mass. However, you are wrong that all neutrons have the same mass because the neutron beta-decays to a proton+electron+neutrino with a half life of about 880 s. So, the uncertainty in the neutron's rest mass energy is about ΔE≈h/(2πΔt)=7.5x10^-19 eV=7.5x10^-28 GeV. The rest mass energy of a neutron is about 1 GeV, so the uncertainty is about 7.5x10^-26 %! This is small but not zero. These results have nothing to do with either the masses of quarks or time quantization.

QUESTION:
Is it correct to say that nuclear fusion violates the law of conservation of mass since a portion of the mass is converted into photons?

ANSWER:
Well, you could say that if there were such a law as conservation of mass. Since 1905 when Albert Einstein showed us that mass is just another form of energy, the only valid such law is conservation of energy. Even in chemistry where conservation of mass appears to be correct, the ultimate source of energy is mass being changed into kinetic energy of the chemical reaction products (heat); chemistry is such an inefficient source of energy that the mass changes are miniscule.

QUESTION:
Can you tell me why muons are extremely unstable (lasting only fractions of a second) while electrons and neutrinos are pretty much stable? I just don't get why muons are so unstable since they're just leptons like electrons and neutrinos just more massive than the other two.

ANSWER:
If you look at the decay of the muon, you will see that the mass of the products, an electron and two neutrinos, is much less than the mass of the muon. This means that the decay is energetically possible, energy is released by the decay so the decay products have kinetic energy afterwards. In nature, almost always when a process is energetically possible it will occur. Only in cases where a decay would be prohibited by some selection rule will decay not occur. For example, a proton cannot decay into three electrons because charge conservation would be violated, even though it is energetically possible.

QUESTION:
I'm writing a research paper for my college english class and the topic is Thorium Reactors. My question is "Are thorium based reactors such as LFTR fusion or fission and based?" I was just wondering because the it seems from what I've learnt that the reactors use the thorium to produce a reaction that makes another element such as uranium 233 which I assume is fusion because I'm sure they're using the energy put off from that initial reaction to power something. But after the uranium 233 is used and to produce energy as efficiently as possible I would think that you would implement a system that would immediately and directly use said produced uranium into some form a fission reactor.

ANSWER:
Fusion always involves light nuclei and there is no such thing as a fusion reactor, only ideas for them. So a thorium reactor must be a fission reactor. It would be inaccurate, though, to call thorium the fuel because thorium is not fissile. If a thermal neutron is absorbed by a fissile nucleus, it will fission and result in more neutrons leading to more fissions and the reaction can be self-sustaining. Thorium, which is 100% ²³²Th, absorbs a neutron to become ²³³Th which β-decays to ²³³Pa (half life 22 minutes)which β-decays to ²³³U (half life 27 days). The ²³³U is fissile and is nearly stable (α-decay half life 160,000 years). Thorium is said to be fertile, absorption of a neutron results in production of a fissile fuel. So a thorium reactor is a breeder reactor, a reactor whose purpose is create fuel. At the startup one needs a starter fissile fuel as well as the thorium to provide neutrons to create the ²³³U. As the ²³³U builds up, it becomes the fuel.

QUESTION:
Why is it necessary to have a minimal temperature of 150 million degrees Kelvin for nuclear fusion on earth if the sun does nuclear fusion at a temperature of 15 milloin degrees?

ANSWER:
There is more than one reason that I can think of. The mass of the sun is 2x10³⁰kg, quite a bit bigger than the mass of fuel in a fusion reactor. Therefore the rate of fusion in the sun can be low but the energy output would still be huge. Increasing the temperature would increase the rate. The density of the sun's core is about 150 g/cm³, 150 times more dense than water. In a fusion reactor, the practical densities are many orders of magnitude below this. Again, the rate of fusion would be dependent on the density of the fuel.

QUESTION:
I've just found out that a proton isn't made up of two up and one down quark but also contains zillions of other up and down quarks along with their anti matter equivalents. Here's a link to an article from the L.H.C people at Cern that shows this. I'm amazed and confused because I thought that matter and anti matter particles would annihilate each other. Why don't they? And, if they do, how is the 'zillions of other quarks' balance maintained?

ANSWER:
The crux of what is going on here is that a vacuum is not really a vacuum as we generally think of it—nothing. Particle-antiparticle pairs are continuously popping into existence and then annihilating back to nothing after a short time. This is called virtual pair production. Also, if a particular particle experiences a particular force, the messenger of that force (gluons for the strong interaction, photons for the electromagnetic interaction) are continuously being emitted and reabsorbed. All this is called vacuum polarization. So, I could give you a similar description of the hydrogen atom: the hydrogen atom is not really a proton and an electron, it is a proton and zillions of electrons and positrons and photons. If you really want to understand the hydrogen atom in detail, you need to take the effects of vacuum polarization into account (see the Lamb shift, for example). The CERN explanation should have included zillions of positrons and electrons and photons inside the nucleus also. Just as a hydrogen atom is pretty well described as a proton and an electron as a first approximation, a proton is pretty well described as three quarks as a first approximation.

QUESTION:
Why the whole matter of radioactive sample does not disintigrate at once or Why it always take half life to disintegrate half of initial value?

ANSWER:
Because decay is a statistical process. Whenever you have a large number N of anything they will, at some time t, have a time rate of change R=dN/dt. If R<0, N is getting smaller (as in radioactivity) and if R>0, N is getting larger (like bacteria growth). For a great many cases in nature, it turns out that the rate is proportional to the number, R∝N. Radioactivity turns out to behave that way: if you measure R for 2x10²⁰ radioactive nuclei it will be twice as big than when you measure R for 10²⁰ nuclei. You can then solve for what R is for a given situation: dN/dt=-λN where λ is the proportionality constant, called the decay constant; the minus sign is put there so that λ will be a positive number since the rate is negative. If you know differential equations, you will find that N=N₀e^-λt where N₀ is the number when t=0. The decay constant is related to the half life τ_½=ln(2)/λ. However, this all depends on there being a large number to start. In the extreme case, if N₀=1, they would all decay at once! You just could not predict when. If N₀=2, they might both go at once or else one might go before the other but not necessarily at τ_½.

QUESTION:
Electrons orbit around the nucleus in varying degrees of proximity to the nucleus. Do electrons farther from the nucleus orbit at a different rate of speed than those closer to the nucleus?

ANSWER:
Yes, if you use a Bohr model for the electrons. However, the notion of electrons running around in well-defined orbits is naïve and incorrect.

FOLLOWUP QUESTION:
Would it be correct to apply Kepler's Laws of Motion to the revolution of electrons about the nucleus in the Electron Cloud Model?

ANSWER:
The electrostatic force is a 1/r² force just like the gravitational force, so if the atom were a classical system Kepler's laws could be used for an atom. In fact, the Bohr-Sommerfeld model, the first extension of the Bohr's circular orbit model, essentially does this by including elliptical orbits and appropriate quantization. Of course, the atom is not a classical system and although such models can be instructive, they are not strictly accurate. What you refer to as the "electron cloud model" would be the proper solutions to the Schrodinger equation, not having well-defined orbits.

QUESTION:
I really don't understand why the detonation of a nuclear weapon does not cause fusion in the gasses in our atmosphere. (Which as I understand was a concern, albeit not a big one in the Manhattan Project.) The detonation of such a device certainly provides more than enough heat and pressure; or so I would think. A little internet searching reveals that thermonuclear devices utilize fission devices to both heat and compress hydrogen to achieve the reaction. So why does one need to add a hydrogen to the fission device to achieve fusion, rather than just rely on the gasses in our atmosphere? Or is it simply that a lot more energy is required to get something like Oxygen or Nitrogen to undergo fusion? All I could come up with thinking of a possible solution was that the shockwave of compressed air these weapons create is made up of gasses that moved away from the source of heat before they could reach that critical temperature, am I on the mark or way off?

ANSWER:
There are many reasons why your ideas will not work.

Fusion requires bringing two charged nuclei close together and the energy required to bring them close enough together to fuse is proportional to the product of their charges. Nitrogen would require 49 and oxygen 64 times the energy to fuse compared with hydrogen.
The thermonuclear device is designed to compress and confine the fusion fuel using the fission device energy to do this. If you just had a fission bomb, the surrounding air molecules would be blown apart from each other rather than compressed.
There is also efficiency to consider. In hydrogen fusion, about 1% of the mass of the fuel is converted to energy but for oxygen or nitrogen, only about 0.1% is.
Finally, you have to ask whether the desired reaction will go at all and what its probability (cross section) is. My guess is that there would be a very low probability for ¹⁴N+¹⁴N—>²⁸Si to occur.

QUESTION:
How much energy is required to move the electron sufficiently far away from the proton such that it does not experience the proton's electric field. I am kind of confused by this question. The effect of electric field will never end as far as we take the electron from the proton. And if we use the relation V=E.L then if E=0, then L will be infinity?

ANSWER:
Technically, yes the field will be exactly zero only at L= ∞. But, that is only if your proton and your electron are the only things in the universe! You can calculate the ionization energy, the energy necessary to move an electron from the ground state of hydrogen to infinity; it is 13.6 eV=2.18x10^-18 J. You can also calculate the energy necessary to move it to any other distance. For example, the energy to move it to about 0.5 μ=5x10^-7 m (which is about 100 times the radius of the hydrogen atom) is 13.599 eV, essentially the same as to infinity. So you can see that most of the energy is supplied in close to the proton because that is where the field (and therefore the force) is strongest.

QUESTION:
So I've been trying to figure this for a while, if a positively charged and a negatively charged quark orbit each other or actually come in contact, also could they form a functional "atom" and be able to have electrons interact with them?

ANSWER:
Two-quark particles are called mesons (one quark plus one antiquark). If you had a positively- and negatively-charged quark you would have an uncharged meson and so it would not bind an electron and you could not have a meson+electron atom. However, if you had a positively charged pion, it could bind an electron. For example, a positive pion π⁺ is composed of an up quark (charge +2e/3) and a down antiquark (charge +e/3) and would have a charge +e and be able to form an "atom" which you might call pionium.

ADDED NOTE:
Whoops! Pionium is an "atom" composed of one π⁺ and one π^-. I cannot find any reference to this pion+electron system, so I guess you can call it whatever you like.

QUESTION:
I read from my history book that when nukes were first created we didn't know that the radiation their explosions produce is harmful to one's health, even to one's life maybe. So my question is, how come neither physicists or medical personnel ever figured out that radiation can be dangerous until we had first victims of radiation?

ANSWER:
Ever hear the expression "hindsight is 20-20"? It is unrealistic to expect that when something new is discovered that we should somehow know all the effects that might have on anything else. At the time radioactivity was discovered, nobody even knew what atoms were composed of or what their structure was. And it was found to be very tiny bits of matter (e.g. what we know as electrons today) and who would have thought that getting hit by something trillions of times lighter than a speck of dust could be harmful? It took experience before it was appreciated how dangerous it could be. One of the best known examples of such experience is the case of radium watch dials. Radium, mixed with a phosphor, was painted by women onto watch dials and it would glow in the dark. The workers were encouraged to point their brushes with their lips to make the fine lines required. Subsequently many became ill with cancer and other radiation sickness. It seems stupid now, but the dangers were not known until they were discovered. Also many of the health effects were long-term effects, the effects not appearing until years or decades after exposure. One example I know about from personal experience is the effect of radiation treatment for acne which was popular in the 1950s. I very much wanted to have this done but my father forbade it, making me furious with him. Years later people who had had this treatment started coming down with cancer thanks, Dad! Yet another example is the shoe-fitting x-ray machine. When I was a kid it was really fun to buy new shoes because you could look down at an x-ray of all the bones in your foot and how well the shoe fitted you; again, when more was learned about radiation, these machines were all sent to the scrap heap. When it began to be appreciated that there were dangers, studies began to be done to try to set allowable dosage levels. But, it is unethical to do such experiments on people, so lab animals had to be used which always presents problems with scaling and other variables. Much of what we know today was learned after-the-fact by doing long-term statistical analyses over decades of medical records.

QUESTION:
Can the equations for Compton Scattering be applied to a photon colliding with a neutron? If no, what equations can be used?

ANSWER:
Yes, you just have to replace the electron mass with the neutron mass. However, the effect for this much larger mass would be very difficult to see because the change in wavelength will be much smaller than for electrons. To have any hope of observing the effect you would need to use extremely high-energy gamma rays.

QUESTION:
I have a question regarding radiation detector. I understand that we have a scintillator (NaI for example) and a photomultiplier tube. Let's say we have an incident gamma ray that produce a photo-electron (forget about x-rays produced by the electrons reconfiguration) in a NaI crystal. How that photo-electron turns again into a photon so it can strike the photomultiplier.

ANSWER:
If γ-rays are being detected, the photoelectron plows through the NaI crystal, ionizing and exciting atoms as it goes. Ideally it gives up all its energy before escaping the crystal. There are now a very large number of photons in visible or IR or UV the number of which is proportional to the energy of the electron which is equal to the energy of the incident γ-ray. Because of reflective coatings on the surface of the crystal, many of these photons reach the photomultiplier tube as a pulse of photons and strike the photocathode which again cause the electron production via the photoelectric effect; because low-energy photons do not cause photoelectric effect, mainly the UV photons participate here. These electrons are then multiplied by the many dynodes in the tube resulting in a current pulse proportional the energy of the original γ-ray. If charged particles are being detected, the charged particle itself plays the role of the initial photoelectron, ionizing and exciting atoms in the crystal.

QUESTION:
What are leptons?

ANSWER:
In the standard model, the fundamental particles fall into two groups, fermions (spin �) and bosons (spin 1). The fermions are divided into two categories, quarks and leptons. There are six quarks (purple in the figure) (u, d, c, s, t, b), the components of hadrons which include neutrons and protons. All quarks have electric charge �1/3 or �2/3 so they feel the electromagnetic force. They also feel the strong force. They also feel the weak force since we know that β-decay changes a neutron/proton to a proton/neutron via the weak interaction. All are fermions. There are six leptons (green in the figure) (electrons (e), muons (μ), and taus (τ), and three associated neutrinos). All feel the weak force. The neutrinos have no electric charge, so they do not feel the electromagnetic force; the charged leptons, of course, do. For each force there is a �field quantum�, that particle which communicates the force between particles which feel that force (the pink in figure 4.8�γ, g, Z⁰, W^�); note that these are all bosons since they have spin 1. For the electromagnetic force, it is the photon. For the strong force, it is the gluon. For the weak force, it is the Z and the W bosons. Note that only the Z has electric charge. The electromagnetic and weak forces have been unified in theory, sometimes collectively called the electroweak force.

QUESTION:
what are electrons,neutrons and protons made up of

ANSWER:
Electrons are believed to be elementary, they have no substructure. Protons and neutrons are made up of quarks.

QUESTION:
I understand that a fusion reaction occurs when two light atoms fuse to form a heavier atom, and some of their mass is converted to energy in accordance with mass-energy equivalence. I am struggling to understand exactly what this has to do with the "binding energy" of a nucleus. When two atoms fuse, some of their mass is converted into binding energy to hold the new nucleus together--but if this is so, where does the energy release come from? If the lost mass becomes energy that holds the nucleus together, it seems that all of the energy would still be contained in the atom. Does the binding energy take only some of the converted mass?

ANSWER:
Your error in logic is assuming that it takes energy to "�hold the nucleus together�". Indeed, if you have a nucleus it takes energy to pull it apart. As an example, the α-particle (⁴₂He) has a binding energy of about 7 MeV per nucleon; a deuteron (²₁H) has a binding energy of about 1 MeV per nucleon. The α-particle has 2 neutrons and 2 protons and the deuteron has one neutron and one proton, so fusing two deuterons should result in one α-particle. The energy released in such a fusion would be (6 MeV/nucleon)x(4 nucleons)=24 MeV. The rest mass energy of an α-particle is about 3700 MeV, so the fraction of mass turned to energy is about 24/3700=0.0065=0.65%. This released energy shows up as kinetic energy of the α-particles�heat, in other words.

QUESTION:
Is there an anti Higgs boson?

ANSWER:

No. There are two kinds of elementary particles, fermions and bosons, and only fermions have antiparticles. For more detail, see this link.

QUESTION:
I'd like to know how counting with half-time, the decaying of radioactive atoms, is possible. I mean if it's impossible to know when atoms will decay, how is it possible to know the half-time decay? I understand the principe but not the possibility. (From a high school student)

ANSWER:

Half life is a statistical property�you need a large number for it to be meaningful. I think you would agree that some radioactive atoms would be more likely than others to decay. With that idea, it should make sense to your that the rate R at which the number of radioactive atoms in a large sample changes will be proportional to the number in the sample, N. So you can write R=-λN where λ is the proportionality constant (called the decay constant) to make the proportionality an equation; the negative sign is there because it is conventional to make λ be positive and I know that if the number is decreasing, the rate of change of N must be negative (getting smaller). I do not know whether you have had calculus or not yet, but to solve for the unknown N(t) you really need it. I will proceed assuming you know calculus: dN/dt=-λN and so N=N₀e^-λtwhere N₀ is the number of atoms when you started measuring (t=0). The half life is defined to be the time when N=N₀/2, so solving, �=e^-λtor ln(�)=-0.693=-λT_� or T_�=0.693/λ. A graph of N(t) for two values of λ is shown to the right. Here N₀=10,000 and you can see that for λ=1 that t≈0.7 when N=5000.

QUESTION:
I just wanted to ask you about your opinion on string theory outrageous claims, things appearing out of nothing, really string theory. String theory math is all too complicated and with no experimental evidence besides the equations. And the hope of putting all that into a"one inch equation" noble but is it possible? I would love to hear your opinion.

ANSWER:

I think you are overreacting a little bit here! My opinion is that string theory is not a theory because it makes no predictions and is therefore not falsifiable or verifiable. However, it is a hypothesis which many good scientists are struggling to develop. And who can predict that any idea is not "possible"? And the math being "too complicated" does not mean that there is no value in the ideas; Einstein found the math needed for the theory of general relativity was "too complicated" for him and he needed to get help from mathematicians.

QUESTION:
Why is Fusion so hard to crack and when can we expect fusion to replace coal and oil?

ANSWER:

Fission is fairly easy because all you have to do is "tickle" a heavy nucleus and it will split in two; this is usually achieved by adding a slow neutron to the nucleus which is easy to do because it has no electrical charge and therefore does not feel any repulsive force from the nucleus. Fusion, however, involves bringing two positively charged light nuclei together. Since they repel each other, they can only get close together if they are going very fast. Another way of saying the same thing is that the fusing target material (generally isotopes of hydrogen) must be very hot. Containing a hot enough gas (plasma, actually, since the high temperature will cause the atoms to be ionized) has turned out to be an extraordinarily difficult engineering problem. There is an old saying among physicists, tongue in cheek, that "fusion is the energy of tomorrow and it always will be!"

QUESTION:
Have you ever heard of internal conversion in radioactive decay (if not I'm fine with that) because a guy on the internet told me that internal conversion happens when atomic nuclei GAIN energy yet sources I read say that it's another way that excited nuclei LOSE energy without emitting gamma rays, is he correct?

ANSWER:
I would be a pretty poor nuclear physicist if I hadn't heard of internal conversion! It is an alternative to γ-decay of an excited nuclear state. The nucleus certainly does not gain energy since the net result (for the nucleus) is a loss of energy as it makes a transition to a lower-energy state. The energy is carried of by the kinetic energy of an atomic electron which is ejected in the process.

QUESTION:
In nuclear fission and fussion there is no change in mass of reactant and product so from where does the 200 MeV energy come from?

ANSWER:
The questioner included the figure above as an example. The total charge (92) and mass number (236) before and after this fission reaction are the same. However, this does not mean that the total mass is unchanged. If you look up all the atomic masses involved in this fission, you will find that the mass is less after the fission. To understand why, see an earlier answer.

QUESTION:
Where do atoms get their energy?

ANSWER:
What energy are you talking about? Of course, an atom has mass energy, mc²; if you just look at a single atom at rest in its ground state, that is all the energy it has. But here is the interesting part: If you take an atom, pull out all its electrons, protons, and neutrons and weigh them, they will weigh more than the atom had before you pulled it apart. So, if all these pieces are at rest and far apart, you have more energy than the atom had. The reason is that you did work to pull it apart and thereby added energy which showed up as mass energy. So your question should really have been "where did atoms lose their energy?"

QUESTION:
What is the difference between muon and neutrino muon?

ANSWER:
A muon is a lepton, sort of like a heavy electron, and is unstable. It has a mass about 200 times greater than an electron. There is no such thing as a neutrino muon, you must mean muon neutrino. Neutrinos are particles with very small mass and are the product of many particle decays. There are three kinds of neutrinos, the electron neutrino, muon neutrino, and tau neutrino and each has its antiparticle, making six in all.

QUESTION:
I am very curious - do different electromagnetic waves/frequencies affect each other in any way? If one photon hits another or if they pass each other, do they affect each other in any way? For instance, if I have a flashlight with a stream of light, and another flashlight with a stream of light shining perpendicularly through the first one, I know that the streams do not seem to affect each other in any way - but do they? In ANY way?

ANSWER:
Indeed, photons interact with each other. However, for all practical purposes, two flashlight beams are not sufficiently intense for there to be an observable rate of interaction. Physicists do study the interaction between two photons, though. One well-known example is the interaction of a high-energy photon with the electric field of a nucleus (and therefore a photon) to create an electron-positron pair.

QUESTION:
What is in a neutron that makes it different from a proton? Is it the stuff that makes up an electron?

ANSWER:
It has nothing to do with electrons. Neutrons and protons, often referred to collectively as nucleons, differ in their internal structure. A proton has two "up" quarks and one "down" quark (leftmost figure) whereas the neutron has two "down" quarks and one "up" quark (rightmost figure). Particles composed of quarks are called hadrons and include also mesons.

QUESTION:
What is radiocarbon?

ANSWER:
I suspect you are asking about radiocarbon dating of organic materials. There are two stable isotopes of carbon, ¹²C and ¹³C. There is one unstable isotope of carbon which exists in nature also, ¹⁴C, sometimes referred to as radiocarbon. Radio carbon is being constantly produced by cosmic rays interacting with ¹⁴N in the atmosphere. ¹⁴C has a half life of about 5730 years, so if cosmic rays suddenly stopped creating it there would be almost none left after a few half lives. But, there is a balance between creation and decay so that the amount of ¹⁴C on earth is about constant at any time. However, suppose you find a fossil which has only about half as much ¹⁴C as all the current living things on earth. Then you can conclude that the fossil is about 5730 years old.

QUESTION:
Elements: Say you have the element hydrogen and it has its electron orbiting at a certain rate of speed and a certain pattern for this entire element. Then you have helium and it has its two orbiting electrons going at a certain rate and pattern, does the two orbit patterns take into consideration each other, in other words does the one electron orbit as if it were alone? Then take the element with three electrons do each electron orbit as if the others weren�t there, or do they affect each other? Then consider an element where the electrons are orbiting as if the pattern was one element but didn�t have the right number of electrons, what element would it be the correct orbiting one or the right numbered one.

ANSWER:
I should first warn you that if you want to understand the details of atomic structure beyond helium, you will have to give up the Bohr-atom, planetary model of the atom. This is a very primative picture of atoms and, while it opened the door to correct quantum mechanical calculations for atoms where we think of orbitals more like clouds of charge-mass around the nucleus, it is essentially an inaccurate representation. That said, the answer to the meat of your question is that any even approximate model of the atoms will include the repulsive forces among all the electrons as well as their attractive forces to the nucleus.

QUESTION:
When nucleons come together from large separation, there is a release of energy called binding energy. Where does this energy come from? If it comes from the mass of the nucleons,then how does the total no. of nucleons remain conserved? If it comes from a part of a nucleon(say proton) then after a part of mass of that proton is lost(as energy) shoudn't we call it anything but 'proton' ?

ANSWER:
If you make a ball out of wood instead of lead, you still call it a ball, don't you? The mass of a nucleon in a bound system is simply less than it is if it is not bound. The reason is simple�E=mc²; in order to pull a bound nucleon out of a nucleus you must do work, i.e. add energy. Where does that energy go? It goes into the mass. In most respects, nucleons in nuclei retain their identities. We know this from nuclear spectroscopy, the energy levels we observe in nuclei. However, don't forget that nucleons are composed of quarks and, to a small extent, the nucleus can behave as a collection of quarks where nucleons may share and exchange quarks. But this is a very small overall effect.

QUESTION:
I was thinking about the particle colliders we have and how how small are the amounts of mass they accelerate. So how small of a fraction of a gram is being accelerated in the particle colliders around the world? Like is a nanogram a good estimate of how much the masses collided total?

ANSWER:
I will use the LHC as an example. The average beam intensity is about 4x10¹⁸ protons per second and the mass of a proton is about 1.7x10^-27 kg. So the mass per unit time would be about 7x10^-9 kg/s=7x10^-6 g/s=7 μg/s.

QUESTION:
How does the quark composition change in beta minus decay/beta plus decay?

ANSWER:
β^- decay essentially changes a neutron into a proton, so the quark composition becomes that of a proton instead of that of a neutron. β⁺ decay essentially changes a proton into a neutron, so the quark composition becomes that of a neutron instead of that of a proton.

QUESTION:
I have been trying to find an answer to a question for some time now. My question is: Can you impart angular momentum to an atom? Whenever I look for information on this I end up with questions of SPIN which when speaking of atoms as you know means something totally different then what I'm asking. I'm wondering if you could "spin" an atom such that its centrifugal force is strong enough to overcome the nuclear bonds of the atom?

ANSWER:
Spin is a kind of angular momentum and certainly not unrelated to what you are asking. But, let's just confine the answer to what you are probably thinking about�orbital angular momentum. Any state of a nucleus has some amount of angular momentum. Just think about a little electron orbit: it certainly has angular momentum. The orbital model is naive, but it gives you the idea that atoms have angular momentum. All even-Z atoms have zero angular momentum in their ground states. There are countless ways you can excite an atom, usually by just shooting something at it, and if you excite it to a state which does not have zero angular momentum, you have imparted angular momentum to it. However, the more angular momentum the atom has, the bigger it is and, as far as I know, it never "flies" apart due to some centrifugal effect; the way to make it fly apart is to give it a lot of energy, not a lot of angular momentum. However, molecules have states which are different from just the orbital motion of the electrons. Imagine a diatomic molecule like O₂, for example. There are states which are rotational states, that is the molecule rotates like a dumbbell. As you give the rotating molecule more and more angular momentum it experiences centrifugal stretching, just like you would expect. If you give it enought angular momentum it will eventually break the molecular bond.

QUESTION:
How can a nucleus have two options to undergo radioactive decay- either by alpha or beta minus decay. ONe process increases the neutron to proton ratio, while the other decreases it. the nucleus must also become unstable by one of the two things- either increase that ratio, or decrease it. then why the two options?

ANSWER:
What matters is the energetics, whether all the mass energy before the decay is greater than all the mass energy after the decay; that way, there will be some energy left over for the kinetic energy of the products. There is an old saying in physics: "Anything that can happen will happen!"

QUESTION:
If an object had zero mass (technically it could not exist) but for the sake of the question let's assume it can, and could withstand any physical affliction and accelerated at 1 km/h would it accelerate at an infinite speed or travel a certain speed only?

ANSWER:
An "object" with zero mass can exist; it is called a photon. However, you must be thinking of this object classically using Newton's second law, a=F/m and so any force will cause an infinite acceleration. In fact the theory of special relativity requires that any object with zero mass can move only at the speed of light in vacuum. So a massless object at rest cannot exist.

QUESTION:
If we see an atom and see all shells and subshells then will I find 3d First or 4s and if we see 4s first then why didn't we name it any other subshell of 3rd shell means why is it 4s? I know that it is chemistry related question but I m confused.

ANSWER:
The letters and numbers mean something. The letters tell you the angular momentum quantum number of electrons in that shell. The letter s tells you that l=0 and 4 tells you that it is the 4th l=0 shell, that there are 1s, 2s, 3s shells with lower energies. The letter p means electrons have l=1, d means l=2, f means l=3, etc. This peculiar labeling of angular momentum quantum numbers is a historical artifact where the words sharp, principal, diffuse, and fine were used to describe spectral lines.

QUESTION:
We did a lab where we put lithium in a flame and saw it emitted red light. We were told this happens because an electron gets so much energy it jumps from one electron shell to the next. Then when it falls back to the lower energy level, it gives off photons of light. We are having trouble understanding what is happening to the electrons. Lithium has two electrons in the first energy level and one in the second. Does an electron in the first shell jump to the second? Or does an electron in the second jump out to the third? And if it is an element with 3 orbitals, does an electron jump from the first to the second, and if it does, does an electron also move from the second to the third? How does the movement of electrons from one shell to the next affect the other electrons in that shell?

ANSWER:
The figure to the left shows the energy-level diagram of lithium. The thing to understand is that the two inner electrons (in the 1s shell) are essentially inert for your experiment, only the outer electron gets excited. So the outer electron (in the 2s shell) looks in and sees the nucleus, charge +3, shielded by two electrons, charge -2, for a net charge of +1. In other words, the lithium spectrum should look a lot like the hydrogen spectrum because the active part of the lithium atom looks pretty much like a hydrogen atom. Note, for comparison, the energy levels for hydrogen shown on the figure. The red line, with a wavelength of about 670 nm, results from the transition from the first excited state (2p) to the ground state (2s).

QUESTION:
I understand you don't answer questions about stars, but im wondering about the nuclear dynamics involved in nuclear fission. In a star with 250 solar masses or more there is something called photofission, where high energy gamma rays cause elements as light as tin to go through nuclear fission. I was wondering, how much energy would be required to cause fission to occur in the element tin?

ANSWER:
Right, this is certainly more nuclear physics than astrophysics, so I can do some rough estimates for you. I can tell you approximately how much energy is released in the symmetric fission Sn¹¹² �>2Mn⁵⁶. The binding energies of the tin and manganese are 953.5 MeV and 489.3 MeV respectively. So the energy released in the fission would be 2x489.3-953.5=25.1 MeV. What I cannot tell you is what energy photon would be required to cause the fission. This depends on the structure of the Sn¹¹² and how it interacts with the photon. Fission might be induced by a much lower energy photon than 25 MeV. After all, uranium can fission with no energy whatever added to it (spontaneous fission).

QUESTION:
What is the difference between atomic physics , nuclear physics , particle physics and high energy physics ?

ANSWER:
Atomic physics studies the atom, mainly the configurations and properties of the electrons which characterize atomic structure; usually atomic physics also includes molecular physics and chemistry may be thought of as applied atomic and molecular physics. Nuclear physics studies the properties of the atomic nucleus. Particle physics studies elementary particles, constituents of atoms and nuclei. High-energy physics usually refers to physics done using high energy particle accelerators, in the GeV range or above, to study mainly nuclear physics or particle physics; can also refer to cosmic ray physics. There is no clear boundary among these areas of physics and there is often considerable overlap.

QUESTION:
Could you explain why an electron in a stable orbit around a nucleus does not emit electromagnetic waves or photons. It only emits when the electron changes from a higher energy orbit. In a stable orbit the electron still is moving around so why doesn't it create a changing E field and radiate? Thanks, Paul

ANSWER:
The glib answer is simply that that is the way nature works. Here we encounter an example of how we should not extrapolate the behavior of something from a regime with which we have experience (for example, accelerating electrons in a transmitting antenna) to regimes where we have no experience (for example, tiny atoms). At the scale of atoms, the wave-like properties of electrons become important and quantum mechanics must replace Newtonian mechanics; when you solve the problem using the correct mechanics, stable orbits occur naturally. I find the easiest way to understand this qualitatively is to note that in an atom a stable orbit is one in which the wavelength of the electron is just right to form a standing wave in its orbit as shown to the right (de Broglie's picture of the Bohr atom).

QUESTION:
My physics lecturer tried his best to answer my question on what exactly a spin in electron is. But I couldn't understand. Besides he himself looked quite confused. So could you kindly help me out? What is generally a spin of an electron? And how would I explain it to a layman?

ANSWER:
Take the earth as an analogy. Earth has two kinds of angular momentum�it revolves around the sun and it rotates on its axis. Angular momentum refers to rotational motion of something. Its motion around the sun is called orbital angular momentum and its spinning on its axis is called intrinsic angular momentum. Now think about an electron in an orbit around a nucleus in an atom. It clearly has orbital angular momentum because of its orbit. But it also has intrinsic angular momentum, that is, it behaves in many ways as if it were spinning on its axis; this is usually referred to simply as spin. This is a fine analogy, but it should be emphasized that spin is not really such a simple classical concept. For example, if you were able to exert a big enough torque on the earth you could stop its spinning or make it spin faster. The spin of an electron is a constant property and cannot be changed. You can change the direction (clockwise or counterclockwise) but not the magnitude of spin. Also, if you try to model the electron as a little spinning sphere to find out how fast it is spinning, you get absurd, unphysical answers. So, you can only push the electron/earth analogy so far. Finally, I should note that in relativistic quantum mechanics (Dirac equation) the existence of electron spin is predicted to be exactly what it is measured to be.

QUESTION:
I know that when electrons get excited they basically move further from the nucleus, but what happens when the nucleus itself goes into an excited state?

ANSWER:
Nuclear structure is more difficult to understand because, unlike an atom, the main force felt by the nucleons (protons and neutrons) is not from some central point but from their nearest neighbors. It was therefore somewhat surprising to find that many aspects of nuclear structure may be understood by considering the nucleons as moving in an average central potential and in well-defined orbits, similar to how electrons in an atom move; this is called the shell model of the nucleus. So one way to excite a nucleus is to simply raise some particular nucleon to a higher orbit as happens in atoms. These are called single-particle states or shell-model states. However, nuclei may also be viewed as fluid drops which behave collectively, that is, all the nucleons move together. Most nuclei are spherical and can be excited simply by causing them to vibrate�imagine a liquid drop which is sloshing around in a rhythmic way. The picture to the left shows the most common shape vibration of nuclei, the quadrupole vibration; the original nucleus (dot-dash line) oscillates between football-shaped (American or rugby) and doorknob-shaped, passing through the spherical shape as it vibrates between the two. Some nuclei are not spherical, usually football shaped, and these nuclei may be excited by being made to rotate.

QUESTION:
Is it true that even an electron has its north and south poles; if it is, then how?

ANSWER:
A simple bar magnet (shown blue) has a magnetic field shaped as shown by the right-most figure to the right. What is fundamental is the shape of the field, not the N and S poles of the bar. This is called a magnetic dipole field. To the left of this is the magnetic field caused by a current loop. Notice that this little current loop has a very similar field shape, so you could identify its N and S poles. An electron has an angular momentum, that is it is spinning. A spinning ball of electric charge is like a stack of tiny current loops, so it will also have a dipole-shaped field. Therefore you could say that an electron has a north and south pole.

QUESTION:
At the LHC protons travel at .999999991 c, or about 3 metres per second slower than the speed of light. I the the proton is also spinning at above 3 metres per second in the direction of travel wouldnt at least part of the proton for a brief period of time exceed the speed of light. I'm guessing the answer is no because all answers involving exceeding the speed of light are no, but i'm having trouble understanding why.

ANSWER:
Many elementary particles, for example the proton in your question, have "spin" which is intrinsic angular momentum. However, you are attempting to model this angular momentum classically, that is by thinking of the proton as a little spinning ball. However, a proton is not a classical object and it is incorrect to model it as such. Any attempt to classically find the speed of the "surface" using reasonable numbers for the radius and mass distribution will result in nonsensical results.

QUESTION:
If I'm not mistaken helium 3 is a very powerful energy source. But there is very little of it on earth. There is lots of it on the moon. so what problems would we run into if we extracted it from the moon why haven't we already done it?

ANSWER:
³He is a potential fuel for a nuclear fusion reactor. The energy producing reaction is ³He+²H �>⁴He+¹H+energy; here ²H is heavy hydrogen or deteurium (proton plus neutron plus electron) and ¹H is normal hydrogen (proton plus electron). There isn't necessarily more ³He on the moon, it is just thought to be more abundant relative to the much more abundant ⁴He there. The interesting thing about helium is that it does not form molecules because it is inert; and, because it is so light, if it appears in the atmosphere it moves so fast that it escapes right into space. The best source of helium on earth is to separate it from natural gas with which it has been trapped underground. The reason that we have not already gotten it from the moon is that it is terribly expensive to send spacecraft to the moon and we would also have to develop the technology to extract it from the moon rocks in which it is embedded and transport it back to earth. Even if that were not a problem, the simple fact is that, in spite of decades of trying, we have been unable to build a nuclear fusion reactor which can produce more energy than it consumes and we are still decades away from having a commercial fusion reactor. There is a kind of funny saying about nuclear fusion: "Nuclear fusion is the energy of the future and always will be!"

QUESTION:
Is there a difference between a proton and photon, besides their difference in charge?

ANSWER:
It is hard to imagine two more different particles:

protons are fermions (spin �), photons are bosons (spin 1)

protons have mass, photons no not

protons are composite particles (quarks and gluons), photons are not

photons are the quanta of the electromagnetic field, protons are baryons

protons have electric charge, photons do not

photons have an "h", protons have an "r"

QUESTION:
As I understand one particle and an antiparticle would, if they come in contact completely, annihilate each other and turn in to energy equaling the mass of both of the particles. But if a particle is moving at close to 1c relative to the observer and the other particle was not moving, the speeding particle would have a relativistic mass. In other words the mass of both particles were not balanced. Would the larger mass of one particle result in more energy, or creation of some other particles? Or would the extra mass from the speeding particle simply transform to normal kinetic energy?

ANSWER:
The questioner used hydrogen-antihydrogen as an example; I have edited this out since those are composite particles and just muddy the water when trying to discuss what could happen. The most common example is the electron-positron pair. If they interact at low speeds the only thing which is possible is the creation of two (or more) photons because there is not enough energy available to create more massive particles. If one (or both) of the pair have significant kinetic energy when they collide, other particles may be created as the questioner suggests. There are selection rules which restrict which kinds of particles may be created, but if there is sufficient energy just about anything allowed could happen�the constraint is that energy must be conserved. In this kind of collision, though, you must also conserve momentum. For example, suppose that you wanted to collide very high-energy positrons with electrons at rest and create a pair of particles with total rest mass M+2m_ewhere m_e is the electron (or positron) rest mass. Your first thought might be that the kinetic energy of the positrons would need to be Mc². But this would not work because the particles would have to be at rest which would violate momentum conservation; so the energy would have to be considerably larger. Proton antiproton interactions are considerably more complicated because they have sufficient rest mass energies to create massive particles even at rest and because they, unlike electrons and positrons, are composite particles made up of quarks. They usually annihilate into mesons which, being unstable, decay ultimately to some combination of gamma rays, electrons, positrons, and neutrinos.

QUESTION:
Outside the nucleus, free neutrons are unstable and have a mean lifetime of 881.5�1.5 s. What is it in the nucleus that keeps the neutrons stable?

ANSWER:
Although somewhat simplistic, you might think of neutrons as stuff inside a nucleus which keeps the protons from getting too close to each other so the Coulomb repulsion will not blow it apart. There is no stable nucleus (apart from ¹H) composed of only protons. On its own, a neutron will beta-decay; quite simply, this is determined by energetics�energy is released if a neutron decays into a proton, a neutrino, and an electron. To find out what happens inside a nucleus, you again look at energetics; if a neutron inside a stable nucleus were to decay, energy would have to be added, so it does not happen.

QUESTION:
Could fusion energy be powered by other hydrogen isotopes or other light elements?

ANSWER:
Yes, fusion produces energy up until you get around iron as the final fusion product. Thereafter it costs energy to fuse nuclei. However, the general trend as you go to heavier nuclei is that the fractional energy output compared to mass gets smaller. Another problem if you are thinking of a reactor is that heavier nuclei have higher electric charge and are therefore harder to bring close enough together to fuse.

QUESTION:
A 5.0 Mev electron makes a head-on elastic collision with a proton initially at rest. Show that: a) The proton recoils with a speed approximately equal to (2E_e/E_p)c and b) the fractional energy transferred from the electron to the proton is (2E_e/E_p), where E_e is the total incident energy of the electron and E_p is the rest energy of the proton.

ANSWER:
I have verified that this is not a homework problem. The solution may be seen here.

QUESTION:
Is there a speed or time measurement for how long it takes the nuclei of Uranium or Plutonium to transfer from Matter to Energy in nuclear fission after the neutron has split the atom?

ANSWER:
I found a paper which lists the "prompt energy release time" as in the range of 10^-20-10^-7 s. This is the time between the scission and the end of prompt gamma rays and neutrons. I presume this huge range is because of dependence on the fissioning nucleus and the fission products.

QUESTION:
"Tritium is produced in nuclear reactors by neutron activation of lithium-6" my question would be; If lithium-7 (the product of the neutron capture) is a stable isotope why does it split? And is this splitting α decay, or fission, that just happens to result in an alpha particle & a tritium nucleus?

ANSWER:
Because the ⁷Li does not get formed in its ground state, rather it is highly excited. Or, you might want to look at it as a nuclear reaction since it happens very quickly:⁶Li+n->⁴He+³H. It is not surprising that this reaction goes because the alpha particle is extraordinarily tightly bound; that is one of the reasons heavy nuclei often undergo alpha decay�because it is such a tightly bound nucleus, there is a reasonable probability that it will spontaneously form inside a nucleus.

QUESTION::
My question is concerning the energy required to pry apart (binding energy) of molecules ( not nuclear). Such as the burning of hydrocarbon fuels. When you do work against gravity, the work (energy) you put in goes into potential energy above the earth. When you pry apart the molecules, you are doing work against the attractive forces holding the molecule together( binding energy)Does that work appear as potential energy (distance) between the molecules ( if the forces remain attractive) OR does the attractive force between the molecules change from attractive to repulsive or vanish with distance? If the forces remain attractive then the work supplied should become potential energy between the molecules. If the forces vanish, or become repulsive, the work (energy) supplied, prying them apart, should manifest as a small increase in mass of the molecules forced apart. I'm not sure how the forces between bound molecules behave with distance as you separate them.

ANSWER:
Molecules are held together by electromagnetic forces, so it is useful to get an order-of-magnitude idea of the binding energy compared to mass energies. Consider a hydrogen atom with an electron and a proton bound together by their electric attraction. The energy necessary to move the electron very far away is 13.6 eV. The rest-mass energy (mc²) of a hydrogen atom is about 1 GeV. Therefore, since you have added energy by ionizing the atom, the atom is lighter by about 100x13.6/10⁹ %≈10^-6%. Any molecular binding energy will be of the same order-of-magnitude. The energy we get from chemistry comes from mass and it is extremely inefficient. So, although E=mc² is at the heart of things, you usually do not have to worry about mass changes in molecular chemistry because they are so tiny. To do detailed calculations of chemical reactions usually requires that you do things quantum mechanically which requires a potential energy function. These calculations can be very complex and approximate models are used to simulate the potential energies of the molecular systems. Once you get beyond the simplest atoms and molecules, the calculations can only be done numerically and approximately on computers. An example of a potential energy function, the Morse potential, for a diatomic molecule is shown in the figure to the left. The form of this potential is V_Morse=D_e[1-exp(-(r-r_e))]²; note that, for one of the atoms in the molecule, the force (slope of the potential energy function) is repulsive for r<r_e and attractive for r>r_e. This is expected since the molecule has a nonzero size because of repulsion but is bound because of attraction. A first approximation often used for bound molecules is a harmonic oscillator potential (masses attached to a spring).

QUESTION:
I know that neutrons is placed in a wax container to contain the neutrons, but why is this an effective way to contain the neutrons. I want to take in account the linear momentum and the energy, but I'm not sure how to begin or where to do the research?

ANSWER:
The wax does not "contain" the neutrons. Most neutron sources yield fast neutrons and most uses for neutrons need slow (called "thermal") neutrons. Thus the source is encased in a block of paraffin which serves as moderator, i.e. the paraffin slows the neutrons down. The reason that paraffin is good is that it is a hydrocarbon and has lots of hydrogen in it. Hydrogen is good to slow down neutrons because the best way to slow down a fast moving object is to collide it with an object at rest which has about the same mass (think of a head-on collision of two billiard balls); hydrogen has about the same mass as a neutron.

QUESTION:
If a particle has a lifetime of say 100 years then what are the chances that that particle will decay in 1 year. How will one go about calculating particle decay probability?

ANSWER:
Lifetime has no meaning for a single particle. It is a statistical concept as I will show. The rate at which a large ensemble of particles decays is proportional to how many particles there are, dN(t)/dt=t/τ where N(t) is the number of particles at some time t and τ is the mean lifetime. The solution of this differential equation is N(t)=N₀exp(-t/τ) where N₀=N(t=0). So, when t=τ, N=0.37N₀, there are 37% of the original particles left. For your question, you might want to ask what is the situation when t=1 if N₀=1 and τ=100: N=exp(-0.01)=0.99. If you want to interpret that as a 1% probablility that the particle has decayed, I guess that that would be ok. But, of course, there is no such thing as 99% of a particle. Keep in mind, though, that this is not linear; for example, if t=50=τ/2, N=exp(-0.5)=0.61, not 0.5.

QUESTION:
What happens to gamma rays when they are blocked by materials like lead, and how come they're not blocked by some other materials?

ANSWER:
For all intents and purposes, gamma rays interact only with electrons in the material. The two most important ways they interact are the photoelectric effect where the photon gives all its energy to an electron and Compton scattering where the photon scatters from an electron, giving some of its energy to the electron. Which of these is more important depends on photon energy and other factors. But, as you would think, the density of electrons in the material is very important. To a very rough approximation, atoms are all about the same size. That means that there are about the same number of lead atoms in a cubic centimenter of lead as there are aluminum atoms in a cubic centimeter of aluminum. But every lead atom has 82 electrons whereas every aluminum atom has 13 electrons. There are therefore roughly 6.3 times more electrons in a cubic centimeter of lead than in a cubic centimeter of aluminum; therefore lead will be much more effective in shielding against gamma rays. Regarding "what happens" to the gamma rays, they disappear completely if photoelectric effect occurs or become gradually less energetic and changed in direction if Compton scattered.

QUESTION:
Two questions regarding radioactive halflife values.
-are these independend of the gravitational field strength?
-are these equal in matter and antimatter?
has there ever been research into this?

ANSWER:
Any clock will be affected by the strength of the local gravitational field (gravitational time dilation); and, certainly, the lifetime of any unstable system is a clock. Your second question must refer to elementary particles since we do not have antiatoms or antinuclei to study. (Actually, antihydrogen has been made in the lab in 2010, but I do not believe there have been any detailed measurements on its properties.) The CPT theorem requires all particle-antiparticle pairs to have identical masses and lifetimes. There has, to my knowledge, never been a case where this is not correct; I am not sure to what extent extremely accurate measurement of antiparticle halflives have been made to test this, but I suspect there is no evidence to suspect CPT violation in this regard.

QUESTION:
In a beta particle emission, do the beta particle and the recoiling nucleus don't move along the same straight line? My friend tells me that. How can be that possible in view of the momentum conservation?

ANSWER:
The reason your friend is right is that there is a third particle emitted in beta decay, the elusive neutrino. Therefore the residual nucleus and beta particle normally do not move along the same line. Historically, since the neutrino is so hard to detect, it was not known that there was a third particle and some people believed that, since there is a spectrum of energies for the observed beta particle, this was a violation of energy conservation. We now understand that the neutrino carries both momentum and energy.

QUESTION:
I was just thinking about the structure of atoms being made of protons, neutrons and electrons. Do you know why atoms even have neutrons if the neutrons are electrically neutral? what is the point of having neutrons? Also, why do they say Pi is squared when it is so obviously round?

ANSWER:
The neutron almost does not exist. A free neutron outside a nucleus is not stable and will decay to a proton, an electron, and a neutrino in about a half hour. However, it is imperative that there be neutrons because there is no nucleus, except ¹H, which consists only of protons. A simple way to look at it is that the nuclear force which protons exert on each other is very strong but, because the protons are so close to each other inside a nucleus, the Coulomb (electrostatic) repulsion is also very strong; so you might think of the neutrons as "buffers" which hold the protons far enough apart that the Coulomb force does not get too big. During nucleosynthesis in a star, if a nucleus is formed which has one or two too many protons, it will decay by transforming excess protons into positrons (anti-electrons), neutrons, and neutrinos.

QUESTION:
Why is the probability of splitting an atom greater when sending slow neutrons rather than fast through the nucleus in fission reactors?

ANSWER:
If a neutron and a fissile nucleus interact, there is a probability, not a certainty, that a fission will occur. A slow neutron will spend much more time interacting with the nucleus and therefore the likelihood of a fission happening is greater than for a fast neutron which spends less time interacting.

QUESTION:
How many particles do the particle accelerators at most accelerate when they are running experiments? One, ten, hundreds or thousands?

ANSWER:
When I was doing experiments some years ago, typical beam currents were on the order of a few nanoamperes. Usually, these were protons which each have a charge of 1.6x10^-19 C. So, suppose the beam current is 16 nA=16x10^-9 C/s; then the number current is [16x10^-9 C/s]/[1.6x10^-19C/proton]=10¹¹ protons/second, about 100 billion. At the LHC in Europe, the intensity is much larger, on the order of 4x10¹⁸ protons/s which would correspond to an average current of about 0.6 A. The high number intensities are vital because the events physicists are looking for are extremely improbable. The LHC is a pulsed machine, its beam is a series of very short pulses. So, the instantaneous current during a pulse is much larger than the average current.

QUESTION:
Is there something wrong with the theory that says, if there is enough dead matter from dying stars that comes together, gravity will take over and eventually produce a new star from this dead matter? I am aware that stars form from hydrogen, but what if 999 trillion tons of the heaviest known elements got together all in one place? Most say a star would not form because there is no hydrogen, but�what then? It would appear that gravity, being indiscriminate, will continue crunching this big iron ball until critical mass is surpassed, regardless of what the result might end up being. What would that result be?

ANSWER:
As I state on the site, I usually do not do astronomy/astrophysics/cosmology questions, so take my answer here with a grain of salt. But, sometimes the question is sufficiently interesting to me that I attempt an answer. I would say that a "normal star" might be defined as an object which makes energy by nuclear fusion, fusing light nuclei to heavier nuclei. However, when you reach iron on the periodic table you no longer gain energy by fusing nuclei and therefore the energy production ceases. Before the whole star becomes iron, though, other instabilities cause the star to, in one way or another, end its life, the details depending on the mass of the particular star. The most dramatic end is a supernova (of which there are several types, again depending on details), a very dramatic explosion of the whole star; the energy from such an event is what is generally thought to be responsible for creation of heavier elements beyond iron. One possible end following a supernova explosion is called a neutron star. The remnants of the star, mostly heavy elements like you refer to, will undergo gravitational collapse and, in essence, electrons will be pushed into protons such that the star is made mostly of neutrons and is extraordinarily dense (it is essentially a giant nucleus). With sufficient mass, that is what I believe would happen to your collection of heavy elements. However, your number, about 10¹⁸ kg, is way too small for the requisite mass. Neutron stars have about 1.4-3.2 solar masses and the mass of the sun is about 2x10³⁰ kg.

QUESTION:
From a physicist's point of view do you ever see us using antimatter as an energy source and are there any ways to even theoretically to create antimatter in adequate quantities? Is antimatter power really just a theoretical thing that won't get any practical use in your opinion?

ANSWER:
So, you have to ask yourself where you are going to get this antimatter. You have to make it because there is never any appreciabale amount just sitting around. You can be sure that it would cost you more energy to create it than you would get from annihilating it.

QUESTION:
if i put a half of Hollow tube insid the water(and the other half is in the air) and drop to it a coin - it will sink. now if i put from the top of the tube a coin that fit the Diameter of the tube and water cant pass this coin what will happen?

ANSWER:
When the coin first touches the water, there will be its own weight Mg down acting down on it, the atmospheric pressure force P_AA acting down on it (A is its area), and the atmospheric pressure force P_AA of the surface of the water acting up on it. Hence the net force will be Mg down and it begins to sink. However, after it has sunk to a depth D the pressure up from the water has increased by an amount ρgD where ρ is the density of the water. So now, Ma=-Mg-P_AA+(P_AA+ρgDA), or a=-g+ρgDA/M. If the water caused no drag on the coin as it fell (obviously not correct), the coin would oscillate about the equilibrium depth D_e=M/(ρA) with simple harmonic motion. In the real world, there would be some drag forces and the coin would eventually come to rest at D_e.

QUESTION:
I want to run a 7.5kw generator for eight hours. I do know it takes 15hp to run a 7.5kw generator. But I want to use weight to do it. Like a grand father clock. How much weight would I need to use if the weight could drop 6ft. ? I would like the equation or equations so I could choose to use more distance or use more weight to make this work for my application.

ANSWER:
So, what is 15 hp in SI units? 15 hp=11.2 kW; this makes sense since some of the power is lost to things like friction. Your falling weight must generate 1.12x10⁴ J/s. If a mass m in kg falls with some constant speed v, the power generated will be P=mgv where g=9.8 m/s² is the acceleration due to gravity. Now, you stipulate that the time must be 8 hr=2.88x10⁴ s, so if it falls h meters in 8 hr, the speed would be v=h/2.88x10⁴=3.47x10^-5h. So, P=mh∙9.8∙3.47x10^-5=1.12x10⁴=mh∙3.4x10^-4; so, your sought equation is mh=3.29x10⁷ kg∙m. So, for h=2 m (approximately 6 ft), m=1.65x10⁷ kg≈16,000 tons. You might want to rethink your plan! If you want to include the time as a variable (in other words, wind your "grandfather's clock" more frequently), the equation would be mh/t=1,140 kg∙m/s. (You will need some kind of governor to regulate the speed and some gearing system to ensure that the translational speed of the mass creates the right rpm for the generator design.) I am afraid that a falling weight is not a very good choice of power source for this application.

QUESTION:
I am confused as to how an anti-particle and a particle can collide. If the fundamental particles only exist at points and don't have any volume, how exactly can two particles ever actually collide with each other?

ANSWER:
Many elementary particles are not point particles. Others may be but the size may be a useless concept on very small scales. But whether or not particles touch begs the question of whether they can interact with each other. All particles have fields which extend away from the particles and it is via these fields that particles interact with each other. This is not solely a quantum concept; an asteroid passing by the earth is deflected even though the two never touch. Similarly, a particle and an antiparticle can interact and annihilate without ever "touching".

QUESTION:
One of my friends (an oncologist, so no dummy) tells me that modern nuclear power plants have a safety switch that will automatically send all the rods into a graphite core in case of emergency which will then prevent a nuclear accident, such as Chernobyl or Fukushima, and will safely contain the fissionable material for.....essentially...ever. I can't find a reference to such a safety device on the internet, although I may be just looking in the wrong place. My understanding is that all nuclear power plants will fail at some point and breach the containment container if water is not constantly supplied. Which is correct?

ANSWER:
There are many kinds of reactors with many kinds of safety systems. What we can learn from earlier accidents is that we should never get cocky about their safety. That doesn't mean that they are not incredibly safe, but surely never perfect. Your friend, though, seems to have something confused. For a nuclear reactor to work, there must be an abundance of thermal (slow) neutrons to initiate the fission. However, each fission releases fast neutrons and, for the geometry of the reactor core, you could not have a self-sustained reaction without slowing them down. The purpose of graphite in a reactor is to slow down the fast neutrons (not all reactors use graphite as a moderator). So to send the fuel rods into graphite would make things worse, not better. The very first reactor, at the University of Chicago during WWII, had bars of graphite gradually added until the reactor went critical (critical means there is one new fission for each fission). The simplest reactor has a bunch of rods called control rods, often made of cadmium, which absorb neutrons efficiently; these are dropped into the reactor core in the event of an emergency to cause the reactor to go subcritical. But accidents can cause damage which cause such safety systems to be inoperative.

FOLLOWUP QUESTION:
Do nuclear power plants require a constant supply of water to remain safe? I'm not talking about the expended fuel rods, but about the core itself. Without water, will a power plant go critical at some point?

ANSWER:
First, the terminology. "Going critical" is not bad, it is necessary for energy to be produced. Here is how it goes:

First with the control rods all the way in, the reactor is subcritical which means that any time a fission happens, fewer than one more occur. Each fission is like a lighted match which goes out without starting a fire.
Now, start pulling the control rods out. Eventually you reach the point where more than one fission happens after each fission; this is called supercritical. So the rate of fission increases and you have to be careful to keep it from "running away".

When it reaches the rate you want, maybe 10²⁰ fissions per second, you have to stop the increasing rate by putting the control rods back just far enough so that the reactor is critical�each fission causes another so now energy is produced at a constant power of P=E∙10²⁰ per second where E is the energy of a single fission. This would be some number like P=1 gigawatt (GW).

Now, to the crux of your question. You are producing energy, in the form of heat, of 1 GW which is one billion joules of energy per second and that energy has to be carried away somehow or else the whole reactor will heat up and melt (the dreaded "meltdown"). The most common way to carry that energy away is with water and the heated water is used to drive turbines to generate electricity. If you do not do something to take the heat away, you have a disaster. (I just made up 10²⁰. Don't quote me on that�it is just to illustrate.)

QUESTION:
Is it possible to accelerate neutrons?

ANSWER:
A scientist is usually loathe to say something is impossible. In order to accelerate something, you must be able to exert a force on it. Does a neutron ever feel a force? Of course it does, the nuclear force or strong interaction is what holds neutrons inside nuclei and this is an extremely strong force. The problem is that this force is also a very short-range force which drops to almost zero as soon as you are a few femptometers (1 fm=10^-15 m) away from the nucleus. So, although a neutron can be accelerated in principle, I can conceive of no way to do it in practice. So, I will go out on a limb and say that it is impossible to build a workable neutron accelerator.

NOTE ADDED:
Neutrons also have a magnetic dipole moment, so they will experience a force if you put them in a nonuniform magnetic field with an extremly strong gradient. Again, this is not really practical for a real-world accelerator.

QUESTION:
what is the difference between l and L in atomic theory?in one book L is given as energy shell and in one book L is represented as orbital angular momentum of electron...so i want to say that energy shell can also be represented as 1,2,3 but orbital angular momentum is always rep.as L so which one is right??

ANSWER:
Notation can be confusing sometimes. In atomic physics, capital letters starting with K are often used to denote shells containing electrons with the same principal quantum number n: n=1 is the K shell, n=2 is the L shell, n=3 is the M shell, etc. The magnitude of the total orbital angular momentum of a quantum state is almost always denoted by L. The value of L is determined by the orbital angular momentum quantum number ℓ: L=ħ√[ℓ(ℓ+1)] where ℓ =0,1,2,3� Also of importance is the projection (component) of the angular momentum on to a z-axis, L_z=mL, where m=-ℓ,-ℓ+1,-ℓ+2�0�ℓ-2,ℓ-1,ℓ. (Note that it is usually conventional to use script ell (ℓ) rather than l.)

QUESTION:
I was wondering, since in quantum phisics observing a phenomenon can change the odds of this phenomenon to arise, how can a detector in a particle accelerator can be trusted? shouldn't it already vary the result simply by measuring them?

ANSWER:
A single measurement causes the probability for that particular particle to be 100% of what you measure. If, however, you make many measurements then you can deduce the distribution of probabilities. One measurement never gets you any useful information in quantum mechanics. For example, suppose you want to measure the mass of a π⁰ meson. This particle decays into two photons, so, if you measure the enegies of the two photons and add them, this is the rest-mass energy (E=mc²) of the meson from which you can easily deduce the mass (m=E/c²). The graph to the right shows the results of many such measurements. Don't worry about the units. The fact is that you might make one measurement which gives you the mass and find that you get that m=0.026 units. You have found that the meson you measured had that mass, but if you keep measuring you will find that others have different masses; you have not caused all mesons to have a mass of 0.026 units just because you determined one to have that mass. You would be inclined to call the mass of the meson to be approximately 0.020 units (the most probable value) with a distribution determined by the rest of your data; the spread of these data, incidentally, would tell you about how long this particle lives, on average, before decaying into two photons. (Oh, by the way, any given meson has the whole distribution of masses before you measure it; measuring it "puts it" at the mass you measure. In other words, a particle does not have a definite mass but a probability distribution of possible masses. Only if the particle is stable (like an electron) does it have a definite mass.)

QUESTION:
While we were studying about atomic structure, our teacher taught us something (I don't remember precisely how was it) about the reason that electron can't be in the nucleus of an atom. She made some calculations and showed that if electron were in the nucleus, it would have to move with a velocity which is greater than that of light. But what about beta decay? Doesn't an electron originate from the nucleus during beta decay? Does it travel faster than light when it originates?

ANSWER:
What your teacher was trying to convey is that an electron cannot be confined in a nucleus, that is, the structure of an atom cannot include electrons which spend most of their time inside the nucleus. There is nothing forbidding an electron from ever being in there. In fact, the wave function of any atomic electron has a nonzero value inside the nucleus so it spends some (tiny amount of) time inside. I am not familiar with her "faster than light" argument; usually the uncertainty principle is used which shows that the energy uncertainty of an electron confined in a box the size of a nucleus is too large to be consistent with measured atomic masses. So, you see, your beta decay example does not violate what she taught you since the electron does not remain inside the nucleus after its creation. (Be sure to realize that the beta decay electron did not exist before the decay happened, it was not hanging around waiting to pop out.)

QUESTION:
I am a high school physics teacher, and am troubled somewhat by inverse beta decay. From what I gather, it can occur when a proton absorbs an electron antineutrino and an up quark flips to a down quark. It then emits a positron and becomes a neutron. What troubles me is where the energy comes from. Flipping a proton to a neutron requires an input of energy, it seems, because of the mass difference. And then there is the mass and kinetic energy of the positron. Can the electron neutrino possibly supply all of this energy? It seems unlikely.

ANSWER:
If you put in the rest mass energies you find that the neutrino must bring in at least 1.8 MeV of energy. This is not a large amount of energy for a neutrino. The main source of neutrinos in the universe is beta decay, so it is to be expected that they would have the requisite energies to induce the inverse processes. For example, neutrino spectra from nuclear reactors are shown in the figure above. As you can see, there are many neutrinos with energies greater than 1.8 MeV.

QUESTION:
In simple artificial bombardment experiments with alpha particles, how do I determine the correct resulting isotope and if/which particles are emitted? For example, 27Al (a,N) 30P results in the release of a neutron. Some such alpha bombardments result in the release of a protons. How do I determine which will occur? Determining either the remaining particle or the resulting isotope is the issue.

ANSWER:
You cannot "determine which will occur." Almost anything will occur as long as the final products are bound. It is just that they will occur with different probabilities. A second consideration is, as you suggest, energetics must be considered. If you subtract all the mass energy after the reaction from the mass energy before the reaction, you get a number called the Q-value. If Q<0, the incident particle must bring in that much energy or else the reaction cannot happen. Here is what we will need: 1 AMU (atomic mass unit)=931.494 MeV/c² and a table of atomic masses (a handy one is here) which gives the atomic masses of isotopes in AMU. I will do a some examples.

Your example: ²⁷₁₃Al(⁴₂He,¹₀n)³⁰₁₅P; M_Alc²=26.981538578x931.494=25133.1413 MeV, M_Hec²=4.00260325413x931.494=3728.4009 MeV, M_n=1.00866491600x931.494=939.5653 MeV, M_Pc²=29.978313753x931.494=27924.6194 MeV. So, Q=27924.6194+939.5653-25133.1413-3728.4009=2.6425 MeV. So, this reaction can occur since there is an excess of of energy after the reaction. There is an additional detail which I will discuss below.
Suppose instead we do the (α,p) reaction: ²⁷₁₃Al(⁴₂He,¹₁H)³⁰₁₄Si; M_Sic²=29.973770136x931.494=27920.3870 MeV, M_Hc²=1.00794x931.494=938.8901 MeV. So, Q=27920.3870+938.8901-3728.4009-25133.1413=-2.2651 MeV. So, for this reaction you must bring in an excess of at least 2.2651 MeV of energy for it to proceed.
Finally, a reaction ((p,3n) which bombards with protons and three neutrons come out) to illustrate my first point above that the final product must be bound: ⁶₃Li(¹₁H,3¹₀n)⁴₄Be. ⁴₄Be would be a nucleus with 4 protons and 0 neutrons and, when you look up its mass you will not find it because it does not exist; 4 protons will not bind together to make a nucleus, ⁶₄Be is the lightest known berylium isotope.

Here is the additional detail mentioned above. The first example, with a positive Q-value might seem to imply that you can just have a bunch of Al and He nuclei at rest and they will react to give you a bunch of neutrons and P nuclei plus a bunch of energy. But, both nuclei are positively charged and they therefore repel each other. Therefore, you must give the alpha particles at least enough energy so the two nuclei can "touch" and interact. I calculate that energy to be roughly 10 MeV for alphas on ²⁷Al, so your reaction will go for the kinetic energy of the α-particles of greater than 10 MeV. For the (α,p) reaction, the kinetic energy must be at least 12.27 MeV.

ADDED DETAIL:
Here is the detail of how I estimated 10 MeV. The potential energy U of two point charges Z₁e and Z₂e separated by a distance R is U=kZ₁ Z₂e²/R where k=9x10⁹J�m²/C² and e=1.6x10^-19 C. The projectile, therefore, must start with its kinetic energy equal to U in order to come from far away to R. I took the size of the ²⁷Al nucleus to be 1.2xA^1/3x10^-15 m=1.2x27^1/3x10^-15 m=3.6x10^-15 m and treated the alpha as a point charge. So, for Z₂=13, Z₁=2, U=26x9x10⁹x(1.6x10^-19)²/3.6x10^-15=1.7x10^-12 J=10.4 MeV. I have used 1 eV=1.6x10^-19 J. This assumes that the mass of the Al is infinite compared to the mass of the alpha-particle, but this is just meant as an order-of-magnitude estimate.

QUESTION:
So I understand that protons and neutrons are made of quarks. A proton is made of two up quarks and one down quark and a neutron is made from two down quarks and one up quark. I know that much. But there are so many other particles in the standard model. Are there any other large particles we can (or cannot) observe besides protons and neutrons that are composed of quarks or bosons or whatever? I have been wondering this for a while.

ANSWER:
There is, of course, no concise answer to your question. The standard model has numerous particles and some are made of quarks (hadrons) and some are not (leptons and field quanta). The hadrons which are bosons (integral spin) are made of two quarks, hadrons which are fermions (half odd integral spins) are made of three quarks. There are also many "particles" called resonances which are excited states of the more fundamental hadrons. The figure to the right which I stole from Wikepedia gives a pretty good overview, I think. For more detail, read the Wikepedia article on the standard model.

QUESTION:
Why any atom having magic numbers(i.e 2,8,20,50 etc.) of nuclides has special stability?(sry for 2 in 1 but i am very curious). How can we determine magic numbers? Is there any specific method to calculate magic numbers?

ANSWER:
Nuclei with particular numbers of protons or neutrons are particularly stable. The origin of these so called magic numbers is very similar to the noble gases of atomic physics where one has a shell model of the structure and a closed shell is particularly stable. In the case of nuclei, it took a lot of work to come up with an explanation for the shell structure. The shell model for nuclei won a Nobel Prize for Eugene Wigner, Maria Mayer, and Hans Jensen. The trick was to recognize the importance of the spin-orbit interaction for nucleons; in atomic physics this effect is also there but very small. The figure to the right shows how the magic numbers appear. On the left are the energy levels predicted for a simple potential well without the spin-orbit splitting. On the right are shown how, if very strong spin-orbit splitting is introduced, new shells emerge. Note that the first three magic numbers, up to 20, are explained without the splitting but not for higher numbers.

QUESTION:
We've learned recently about the strong and weak nuclear forces. We also learned that when an isotope has too few neutrons, the nucleus stabilizes itself by having an electron from one of its orbitals crash down into a proton and turn it into a neutron. What I don't understand is how does the nucleus as a whole know it is not stable, if it just an inanimate collection of matter? In other words, how does a proton know it needs to turn into a neutron, how does it know that there isn't that extra neutron on the other side of the structure, how does it know that it needs to sacrifice if it is inanimate, if it can't see or talk or anything? And furthermore, when this happens, how do the electrons know they have to crash into the nucleus? How do they know where to go, and where do they get the momentum to do this? Who pushes them? And to further my question, who's to say that an isotope with two few neutrons doesn't have all of its protons simultaneously trying to stabilize, causing the whole structure to be a bundle of neutrons?

ANSWER:
This is not exactly a single, well-focused question as stipulated by the the groundrules of this site, but I will answer some of your questions with a general description of beta decay which is what you are talking about. In some sense, that inanimate collection of matter as you call it, does know what to do because it is a quantum mechanical system subject to rules. The most important rule, perhaps, is that a nucleus will seek the lowest energy state. Consider a nucleus which has an atomic weight A, composed of N neutrons and Z protons such that N+Z=A. Nuclei of a given A are called isobars. So, N could be anything from 0 to A with Z=A-N. In practice, nuclei which are bound at all must have both protons and neutrons. Take a specific example: Plotted to the right is the binding energy of the known isobars of A=73 plotted as a function of Z. Binding energy is essentially the energy you would have to supply to totally disassemble a nucleus and the more tightly bound a nucleus is, the lower the energy state. So, if nature provides a way to keep A constant but change Z and N, a nucleus will take advantage of it. So, Zn and Ga have too few protons (or too many neutrons, however you choose to look at it), and they will decay by turning a neutron into a proton plus an electron plus a third particle called a neutrino (which has practically no mass and no charge). Because the energy of the nucleus decreases, the leftover energy is used to shoot out the electron and neutrino. This is called β^- decay. But, when these two get down to Ge they will go no farther because energy would have to be added to get to As and there would be none left for the electron and neutrino. On the other side, As and Se have too many protons (too few neutrons) compared to Ge and so they have to hope that nature provides a way to turn protons into neutrons. Indeed, a proton can decay into a neutron plus a positron (a positively charged electron) plus a neutrino. This is called β⁺ decay. Finally, there is a process which can compete with β⁺ decay which is called electron capture which is what you refer to in your question. There is a swarm of electrons around the nucleus and if a proton could combine with one of them it would do so. Indeed, this is allowed if a neutrino is ejected. The electrons do not have to suddenly "crash into the nucleus", they are already there. The electrons in an atom do not really run around in little circular orbits, they are smeared everywhere, even inside the nucleus; there is a small but nonzero probability of looking for an electron inside the nucleus and finding it there. I think if you read through my answer carefully you will find I answered all your questions. Nature is "smarter" than you give her credit for!

QUESTION:
We've been learning about nuclear decay and radiation lately in AP Chemistry( I am junior in high school), and this question crossed my mind: If protons can absorb electrons, what if a neutron absorbed a positron? I was imagining that if we had a neutron emitter pointed to the same place a positron emitter was pointed, positrons might merge into some neutrons and create protons. Then i wondered, now what if we were to isolate this new proton into a separate chamber of this weird apparatus I cannot fully explain, in which we emitted electrons, and thus making the proton into a neutron again, except it is now one electron and one positron heavier. Now what if we continued this cycle? Would the neutron become unstable as do heavy elements? Are current-day neutrons nothing but the development of the said process to the most stable form, as the amount of positrons and electrons must be just right? If we managed to force more positrons and electrons into the neutron than naturally allowed, would it be like nuclear fission, but make an explosion that is thousands of times greater than a hydrogen(fusion) bomb? Or will there be such a tiny mass defect when the neutron gets too large that it only explodes the apparatus and nothing more?

ANSWER:
Here is the thing to appreciate first off: a neutron is a neutron is a neutron. There is no such thing as a heavier or lighter neutron, its mass must be equal to the known neutron mass. Likewise for protons or any other elementary particle. So, most of your question has no answer because the basic premise is wrong, but I will tell you a bit about these processes (generally called beta decay or weak interaction processes). No matter how you choose to create any elementary particle, the energies (including the mc² energies of the particles) of all the participating particles before and after the creation must be exactly the same. Electron (e-) capture by a proton is a common nuclear decay mode, resulting in a neutron (n) afterwards. Similarly, positron (e+) capture by a neutron is possible (but rare) and results in a proton (p) afterwards. One thing you miss, though, is that there is a third participant, a neutrino (ν), which exits afterwards in both cases. The reason that electron capture is common is that the nucleus is surrounded by a cloud of electrons available for capture; positrons are not normally around unless you put them there and then they are much more likely to annihilate with one of the many electrons before getting the chance to interact with a neutron in the nucleus. So, let's talk about the energetics of these reactions:

Rather than using masses in kg, I will use rest mass energies Mc² in MeV (million electron volts). [I assume that, since you are in an AP course and talking about nuclear chemistry, you are familiar with the electron volt as a measure of energy; if not, it is the energy an electron acquires when accelerated across a potential difference of 1 volt.] The relevant masses are: M_nc²=939.565378, M_pc²=938.272046, and M_e-c²=M_e+c²=0.510999. The neutrino has approximately zero mass. Although these are energies, it is customary, if a bit sloppy, to refer to these numbers as masses which I will do.
We will also need sums of masses, M_pc²+M_e-c²=938.783045 and M_nc²+M_e+c²=940.076377.
For electron capture, p+e^-―> n+ν. Note that the sum of the mass on the left is 938.783045 and on the right is 939.565378. That means that if all masses are at rest before and after the capture, this is impossible because there is not enough energy before to create a neutron. The way this can happen is if the electron brings at least 0.782333 MeV of kinetic energy with it when it encounters the proton and then any leftover energy will be shared by the kinetic energies of the neutron and the neutrino. The energetics also tell us that the neutron is not a stable particle because its mass is greater than the masses of the particles it could decay to and a free neutron will decay, n―> p+e^-+ν. The half life of a free neutron is about 15 min. In a stable nucleus a neutron does not decay.
For positron capture, n+e⁺―> p+ν. Note that the sum of the mass on the left is 940.076377 and on the right is 938.272046. There is adequate mass to create the new mass and the excess energy is carried off by the neutrino. It also tells you that the free proton is a stable particle (at least against decaying into a neutron and positron) because where would the extra energy come from?

So, a proton (neutron) does not get a little heavier each time it combines with an electron (positron) and then with a positron (electron), you end up with the mass of the proton (neutron) again.

QUESTION:
1) is it possible to know the exact location of a single atom? 2) If we use the most powerful microscope in the world and make it 1 million times more powerful would we be able to physically and directly see what is happening at an atomic level? would we see the atoms?

ANSWER:
It is not possible to know the exact location of anything because of the uncertainty principle. However, it is certainly possible to see individual atoms but not, as you suggest, by making a more powerful optical microscope. A conventional microscope will not work since once you get on the scale of the wavelength of the light you are using you begin to get unavoidably fuzzy images and eventually no image at all. Atoms are usually detected individually by something called th atomic force microscope.

QUESTION:
Is the Proton Stable, or does it decay over time? For instance I read it's Mean lifetime is: >1.9 x 10^29 years (stable) it confused me because of the stable in the brackets, is it saying that it is stable and that the lifetime is actually hypothetical?

ANSWER:
Although the proton is predicted by the standard model to be stable, there are other theories which predict it to decay. It is therefore of interest to try to observe proton decay experimentally. Experiments have all had negative results. However, a negative result of an experiment usually does not mean that the sought result does not exist, it puts an lower limit on its happening. For example, the analysis of an experiment might result in being able to say that the proton is stable unless its lifetime is longer than, say, 10³⁵ years. A later more sensitive experiment might be able to lower the limit to maybe 10³⁰ years, and so on.

QUESTION:
what is the differences between the meaning of charge and matter densities of a nucleus?

ANSWER:
The charge density distribution is a description of how electric charge is distributed in the nucleus. The matter density distribution is a description of how the mass of the nucleus is distributed. These turn out to have roughly the same shapes for nuclei, that is the radii are about the same, both are more or less constant inside the surface, and the surface is reasonably well defined, that is the distributions drop to zero in a distance small compared to the radii. This need not be the case because the charge is determined by the protons whereas the mass is determined by both the protons and neutrons. There is evidence that, in some nuclei, the mass distribution is slightly larger than the charge distribution; the mass outside the charge distribution is sometimes called a "neutron halo".

QUESTION:
Particle accelerators are said to reproduce the early conditions of the universe. At the same time the energies of these collision are said to be no greater than that of cosmic rays hitting the upper atmosphere. Can we say then that the early conditions of the universe are constantly reproduced in our own upper atmosphere? Do you think it would be feasible in the future to just build a detector in the upper atmosphere in the future? Can't we just point a very sensitive telescope toward our skies and learn from these collisions?

ANSWER:
Yes, some cosmic rays have extraordinarily large energies. But to study the physics you need two things: since the events you want to study are very rare, you need a very intense beam of particles to hope to see anything of interest in your lifetime; accelerators give large numbers of particles per second striking a very small area, cosmic rays do not. Second, one needs control over the conditions. Cosmic rays can have a broad spectrum of energies, an unpredictable location where they will hit, and an uncertainty as to what they will interact with if anything at all; accelerators put the particles where you want them, with the energy you want them to have, and on specifically what you want them to hit.

QUESTION:
Conversion of mass to energy (fission) has been demonstrated many times in laboratory and field tests. Has conversion of energy to mass also been demonstrated in laboratories?

ANSWER:
Yes. A couple of examples:

A very energetic photon (massless) can spontaneously turn into an electron-positron pair; this is called pair production.
The mass of a nucleus is always less than the sum of all the constituent proton and neutron masses. Suppose you remove a neutron from a nucleus; it will take work because that neutron is bound in the nucleus. Hence, the final system of neutron and the original nucleus minus one neutron has a greater mass if both objects are at rest. So let�s just say that the nucleus, having a mass smaller than the sum of its parts, is an example of converting energy into mass because there is more mass after you disassemble it by adding energy (doing work).

QUESTION:
Rutherford was able to determine the nature of alpha particles by recognizing that they have the same atomic spectrum as helium. However, if alpha particles are just the nuclei of helium atoms, then in the absence of electrons, what produced the spectra?

ANSWER:
He put his alpha source in a vacuum and waited a while. Alpha particles would acquire the necessary electrons and, when there were enough of them, he could do the requisite spectroscopy.

QUESTION:
Why is radioactive decay dependent on the number of atoms or radioactive particles present in the sample? Does it have anything to do with the total energy of all the atoms in the sample?

ANSWER:
For every radioactive nucleus there is a probability that it will decay in a certain elapsed time. If you have a thousand of them, then you will see them decay at a rate which depends on the probability tha any one will decay. If you have a hundred thousand of them, the rate at which they will decay will be 100 times the rate at which you observe the thousand to decay. Given the fact that the likelihood of any given nucleus decaying is the same as for all the others, then the number you actually see decay per unit time is proportional to the number of nuclei present.

QUESTION:
I'm an English Ph.D. student writing about the figure of the microcosm in the modern novel, and am interested in the parallel with contemporary physics and the conception of the atom as a "little solar system." My question is: how much did the evidence from the experiments done by Rutherford, et. al. make this theory necessary and how much was it just a convenient metaphor? Would other pictures of the atom have been equally plausible, or was this pretty much as accurate a representation as possible?

ANSWER:
As I have said in many earlier answers, the Borh model of the atom, electrons running around in well-defined circular orbits like planets around the sun is, at best, a very rough approximation of atomic structure. Historically, such a model was ruled out because an electron moving in a circle radiates its energy away and it would very quickly spiral into the center of the atom. However, Rutherford's experiments showed unambiguously that nearly all the mass of an atom and all its positive electric charge resided in a volume extraordinarily tiny relative to the size of the atom. Putting the electrons outside then became imperative and Bohr invented his model of the atom by postulating certain rules for orbits which would be special in that electrons in those orbits would not radiate away energy. His rules also had to, and did, explain the spectrum of the hydrogen atom, those colors of light which are emitted from hydrogen. But, this is clearly ad hoc in many ways as new ideas often are. Within just a few years, though, partly motivated by the success of this model, a new branch of physics, quantum mechanics, was developed which provides an accurate picture of both how we should think about atomic structure and why certain atomic states do not radiate away their energy.

QUESTION:
I have been told that the strong force becomes repulsive at small distances. Is this the case and can you explain why or why not?

ANSWER:
It is certainly true. I cannot explain why since that is not really the goal of physics; we don't, for example, ask for an explanation of why the electron is negative, it just is. The veracity of the repulsive short-range force is easy to understand. If it were not so, the nucleus (held together by the strong interaction) would collapse. This is called saturation of nuclear forces.

QUESTION:
why does CO2 leak very quickly from a bicycle tube, while N2 (rather than pesky messy air) leaks very slowly? I'm told "no easy answers", but maybe there is for the specific situation?

ANSWER:
Actually, there is an easy answer. CO₂ is a much smaller molecule than N₂.

FOLLOWUP QUESTION:
can you elaborate just a bit? for this 7th grader mentality, it would seem that N2 & O2 should be about the same size - aren't all electrons are in the same orbital shells? and then the addition of a C to the O2 should make it bigger? is it something to do with the double bonds on the CO2 that makes the molecule smaller?

ANSWER:
You should not think of atoms as little balls of constant size. What really determines how large a molecule is is how strongly bound it is. Imagine atoms A and B make a molecule AB and C and D make a molecule CD and that all four atoms have about the same sizes. If A and B attract each other much more strongly than C and D (i.e. A and B are more tightly bound), then we can expect AB to be smaller than CD. I am not a chemist, but maybe your double bond idea is a way of saying this. I am a nuclear physicist and there is a similar effect in the following example: a deuteron, consisting of a proton and a neutron has a radius of about 2.1x10^-15 m and an alpha particle, consisting of two protons and two neutrons, has a radius of about 1.6x10^-15 m. The alpha particle is much more tightly bound than the deuteron.

QUESTION:
What determines the number of neutrons in an atom?

ANSWER:
The number of protons in an atom of a given element is fixed (atomic number). However, the number of neutrons might vary and nuclei of the same element with different neutron numbers are called isotopes. There are often two or more stable isotopes for an element. The number of neutrons is determined by the nuclear structure, that is by the forces between neutrons and protons in a nucleus, and is too complicated to try to explain here. The lightest nuclei tend to have equal numbers of neutrons and protons but, as the atomic number increases beyond about 20 or so, there tend to be relatively more neutrons. For example, one stable lead isotope has 82 protons and 126 neutrons.

QUESTION:
I know a little about the bohr model of the atom and I am curious about how the nuclear charge of an atom affects it's radii. If the nuclear charge increases I would think that the radii would decrease but I'm not sure of the math and if that model holds up.

ANSWER:
If you do the Bohr model except for a central charge of Ze instead of e (i.e. magnitude of the force on the electron is k_eZe²/r², you will find that the radius is a₀/Z where a₀ is the radius of the hydrogen orbit in the ground state.

QUESTION:
The P in PET scan stands for positron. Positrons are antimatter. How are the positrons transported and injected without destroying the instruments and how is the positron selecting the correct tissue to destroy?

ANSWER:
The patient is not bombarded by positrons. What happens is that she is injected with radioactive nuclei which emit positrons. The positrons, very close to where they decay, encounter an electron and they annihilate resulting in two photons which are detected and their trajectories traced back to where the annihilation took place. Read the Wikepedia entry on PET.

QUESTION:
How do you make something radioactive? Can you simply hold it next to a radioactive item?

ANSWER:
Holding something close to a radioactive source will not cause it to become radioactive. Suppose you start with something stable (as most naturally occuring things are) and you want to make it radioactive. You have to change the nucleus to one which is unstable. The most common way is to expose the sample to slow neutrons. The reason is that neutrons have no electric charge and so they are not repelled from the nucleus and, if they move slowly, they take a relatively long time to pass through and therefore have a relatively high probability of being captured. Usually the sample to be activated is put into a nuclear reactor where there are copious amounts of neutrons. One example would be ⁶⁰Co, a radioactive isotope of cobalt which is commonly used for cancer treatment. Stable cobalt which is composed of ⁵⁹Co absorbs a neutron and becomes ⁶⁰Co which has a half life of about five years.

QUESTION:
why don't electrons in an atom collide with the nucleus? Doesn't the nucleus of an atom have a positive charge? if so, then it should make sense that the electrons would be attracted to it, and not just go around it in orbitals. So yeah, I'd like to know why electrons don't fly into the nucleus, and stay in their orbitals. At the quantum level, I doubt gravity has any role in it, and even if it does, the mass of the nucleus would far succeed the mass of any individual electrons.

QUESTION:
What keeps an atom from getting smaller and smaller if an electron can lose energy and be attracted to the nucleus? What keeps electrons from moving closer and closer to the nucleus?

ANSWER:
The electrons are attracted to the nucleus, that is the force that holds them in their orbits. Your question is akin to "why don't the planets in the solar system collide with the sun?" Quantum mechanics shows that there is a lowest energy a system can have, called the ground state, and for atoms this is always above zero and if an electron "popped into" a nucleus and just sat there this would be a state with lower than the minimum allowed energy. On the other hand, the idea of orbits for atoms is handy but not really very good. The electrons should be thought of as a cloud and the density of the cloud at any location is representative of the probability of finding the electron there. This cloud extends all the way to the center and so there is a small but nonzero chance of finding the electron inside the nucleus. The picture at the right shows this cloud for the ground state of a hydrogen atom. The densest part is where the Bohr orbit would be.

QUESTION:
While reading " A brief history of time by stephen Hawkings, i found that he explained something about spin 0,1,2. I wanna know that how is spin 0,1,2 different from Spin 1/2 and how does pauli exclusion principle explains it?

ANSWER:
All these numbers are spin quantum numbers. If the spin quantum number of a particle is s, then the intrinsic angular momentum of that particle is S={√[s(s+1)]}h/(2π). Intrinsic means that this is not due to some orbital motion, but is associated with even a free particle. It is sort of like the earth whose orbital motion around the sun and rotational motion about its axis are different things; particle spin is sort of like the earth's rotational motion about its axis. Particles with half odd-integer spin quantum numbers (1/2, 3/2, etc.) are called fermions and obey the Pauli exclusion principle; examples are electrons, protons, and neutrons (all with s=�). Particles with integer spin quantum numbers are called bosons and do not obey the Pauli exclusion principle; examples are the pi meson (s=0), photon (s=1), and the graviton (if it exists, expected to have s=2).

QUESTION:
Do protons generate magnetic field around them ?

ANSWER:
Any electric charge which is moving creates a magnetic field. Since a proton has charage, it will generate a magnetic field when moving. However, a proton also has a magnetic moment, that is it looks like a tiny bar magnet. Therefore a proton has a magnetic field even if it is sitting still.

QUESTION:
As I uderstand it, Carbon 14 is formed by neutrons comming from cosmic rays witch interract with Nitrogen(7 protons + 7 neutrons). Nitrogen lose a proton and win a neutron wich become carbon 14 (6 protons + 8 neutrons), right ? Then, the part that I do not get is how Carbon 14 decay to Nitrogen. Wikipedia says that Carbon 14 loses eletrons and electrons antineutrinos to become Nitrogen again, but I don't understand because Nitrogen is 7 protons. Is Carbon gaining a proton ? Can you make this more clear for me ?

ANSWER:
The ¹⁴C decay is what is called beta decay. Inside the ¹⁴C nucleus one of the neutrons spontaneously turns into a proton, an electron, and an antineutrino. The electron and the antineutrino leave and the net effect has been to change ¹⁴C to ¹⁴N.

QUESTION:
How much energy does CERN use to smash 2 protons together to try & see the Higgs particle? ie in laymans terms, would it be equal to lifting a chevy off the ground...or a train engine?

ANSWER:
There is a link from CERN which fully discusses the energy content of the LHC beams. The number they give for the beam at full intensity is 362 MJ. They show that this is the energy of a 3200 kg Subaru going about 1000 mph. It is the energy which could lift that car about 10,000 m, or about 30,000 feet. But maybe you are asking about just the energy of the two protons? That is much smaller, because the beam contains about 3x10¹⁴ protons and so the energy of each proton would be about 10^-6 J, not enough to notice macroscopically.

QUESTION:
What would happen if one split a radium molecule? This question is from my ten year old daughter's who has been reading about/admiring Marie Curie for the past two years.

ANSWER:
I think you must mean split a radium nucleus. Indeed, if you were to induce a fission of radium, energy would be released just like it is from any other fission of a heavy nucleus (like ²³⁵U or ²³⁹Pu), but you could not make a reactor with radium as a fuel because it is not fissile which means that it can be induced to fission by causing it to absorb slow neutrons and to therefore sustain a chain reaction. The energetics are such that if it were split into two fragments, energy would certainly be released. You can tell your daughter that I had the pleasure of working with Marie Curie's grandaughter Mme. H�l�ne Langevin-Joliot, also a nuclear physicist, at Saclay National Laboratory outside Paris in the 1990s.

QUESTION:
Is there any relationship between Lennard-Jones potential and Van der Waals force ?

ANSWER:
Yes. The Van der Waals force is the long range attraction between neutral atoms or molecules. The Lennard-Jones potential represents an empirical force which has both a short range repulsion (r^-12) and long range attraction (-r^-6); the attraction part could be called Van der Waals.

QUESTION:
what is the spin orbit force?

ANSWER:
An elementary particle may have two different kinds of angular momentum, spin and orbital. For example, if you think of an electron of mass m in an orbit of radius r and speed v around the nucleus, it has orbital angular momentum which classically is mvr and quantum mechanically is ħ√[ℓ(ℓ+1)] where ℓ is an integer. But, it also has a spin angular momentum of ħ√[�(�+1)]. Now, if there is some way these two separate angular momenta can interact with each other, the result will be that the energy of the electron will depend on the relative orientations of the angular momentum vectors, either parallel or antiparallel. This is called spin-orbit splitting or fine structure in atomic physics. In atomic physics it is pretty easy to understand why the angular momenta couple. From the electron's point of view, it sees the nucleus orbiting it and therefore sees a magnetic field due to that current; the electron has an intrinsic magnetic moment (it looks like a tiny bar magnet) antiparallel to its spin which wants to align with that field which means that it takes work to move the electron to the unaligned orientation meaning aligned is at a lower energy. The spin-orbit interaction is a very much bigger effect in the structure of the nucleus but its origin is not as easy to understand as for electons in atoms.

QUESTION:
A stable compound nucleus has a lower mass than the sum of its constituent protons and neutrons. I was told this is because the bound state has lower energy. A similar situation occurs in the atom: the stable bound state of an atom has a lower energy than its constituents, and thus the mass of the atom is slightly less than that of the electrons and protons that make it up. This leads me to believe that a bound system minimizes energy and hence mass. Why does this reasoning fail when thinking about the binding of quarks? The up and down quark masses are around 1-6 MeV, but the bound states of quarks, rather than being less than the sum of their constituents, are much larger (pion ~130 MeV, proton/neutron ~1 GeV). It seems that it is not necessary for a stable bound state to minimize energy. This seems strange to me. Do you see where my confusion is coming from?

ANSWER:
You are sort of muddled here. Here are the basics:

Any bound state is less massive than the sum of its parts; this is easy to convince yourself of because it obviously takes work to disassemble any bound state and therefore energy must be added. This energy (which you put in) shows up as increased mass.
By definition, the ground state is the "boundest" state of a system. Therefore the ground state is the least massive.
Conversely, if a system binds, the resulting object has less mass than what you started with and therefore energy is released. For example, CO₂ is less massive than C+O₂ and that is where the energy comes from when you burn carbon�fossil fuels, essentially.
Because of the way quarks bind, there is no such thing as a free quark and therefore it is difficult to talk about its rest mass. To zeroth approximation, it takes infinite energy to unbind a quark so it would have infinite mass.

QUESTION:
if 50 billion neutrinos pass through your body every second .... I have read this so many times .... why do only a few show up in the detectors below ground. I thought that neutrinos could travel through anything.

ANSWER:
What are you asking? If they are very unlikely to interact with your body, they would also be very unlikely to interact with the underground detectors. It is really hard to detect neutrinos and that is why the the detectors are so huge (millions of gallons)�to enhance the chance of "catching" one.

QUESTION:
I am having trouble comprehending the idea of a neutrino having no mass. Wouldn't that mean that it simply does not exist? How does something exist that does not exist? Do we just not have the technology to measure it's mass?

ANSWER:
First, let's address the question of the "existence" of massless particles. Possibly the most abundent particles in the universe, photons, are massless. Photons are particles of light, and there is no argument that they exist. The restriction which must be obeyed by massless particles is that they must, in a vacuum, travel with the speed of light. You can never find a photon at rest. So other massless particles might exist. For many years neutrinos were thought to be the other known massless particle in our universe. However, recent experiments have shown that neutrinos cannot be massless, but their masses (there three kinds of neutrinos) are known to be nonzero. However, they are exceedingly tiny, hundreds, maybe thousands of times smaller than the mass of an electron. Therefore most neutrinos travel with a speed very, very close to the speed of light.

QUESTION:
I was wondering. Just as one can record a video in 5000 fps, how many equivilant "fps" does the Atlas-detector at LHC record at? If that comparison is even possible?

ANSWER:
The protons travel in bunches. The bunches collide at a rate of about 40 MHz=4x10⁷ per second. Each bunch collision results in about 25 potentially interesting proton-proton collisions per second. So, the average event rate is about 10⁹ per second. The detector can store about 200 events/second, so very complex and elaborate triggoring systems must be used to distinguish the interesting from less interesting events.

QUESTION:
What's stopping entrepreneurs from using fission to turn mercury or other elements into gold or silver? Do you think there will come a day when we simply can create whatever elements we want? Any guesses as to how far in the future that would be?

ANSWER:
The issue is the rate at which you can create something. The day has already come when we can create whatever elements we want, but can we create them in usable quantities? Fission is not what you should be thinking about, because only a few elements (uranium and plutonium) can sustain a chain reaction. But, using particle accelerators or neutrons from reactors, elements may be made to transmute. So, you need a very intense beam of particles to bombard with and a reaction with a reasonably high probability of happening. Suppose you were able to find a reaction where you could cause a million reactions leading to gold per second. But how many gold atoms do you need to get a macroscopic amount? Avagadro's number is on the order of 10²⁴. To get 10²⁴ atoms would require a time of 10²⁴/10⁶=10¹⁸ s≈30 billion years.

QUESTION:
If my understanding is correct, Iodine-131 is a product of the fission reaction of uranium-235. If this is correct, then by the conservation of proton number, the other product has to be an isotope Yttrium. However this means the isotope has to have a crazy mass number or a very large number of neutrons are released. So my dilemma comes down to understanding how to complete the equation: U-235 + 1n ---> I-131 + ? + ?neutrons.

ANSWER:
Here is what happens in nuclear fission. First the ²³⁵U splits into two nuclei which are extremely neutron rich. The two are so neutron rich that they almost immediately eject some neutrons, typically 5-8 neutrons between them. But, after this, the remaining two nuclei are still very neutron rich; therefore, they both undergo a series of β decays where each decay results in a neutron turning into a proton plus an ejected electron and neutrino. So, you can see, you cannot tell what the initial decay products were just by looking at the final decay products because the β decays result in a lot of negative electric charge leaving the system.

QUESTION:
What about the structure of alpha particles makes them such a common form of radiation? Why aren't, for instance, lithium nuclei the common form of ionizing radiation?

ANSWER:
Alpha radiation is something which happens mostly for very heavy nuclei. Think of all the protons and neutrons buzzing around inside the nucleus, all interacting with each other. It turns out that the alpha particle (⁴He, 2 protons and 2 neutrons) is the most tightly bound of the light nuclei. The figure to the right shows the binding energy per nucleon (the average energy it takes to remove a nucleon from the nucleus, E/A, where E is the energy to totally disassemble the nucleus and A is the atomic number) of stable nuclei. So the alpha is tightly bound and has few particles and the result is that it has a relatively high probability of spontaneously forming inside the nucleus. The energetics are such that the total energy of the original nucleus ^AX^Z_Ns greater than the energy of an alpha particle plus the daughter nucleus ^A-4W^Z-2_N-2and so, alpha particle decay will happen if a mechanism for decay can be found. It gets a little complicated, here because the alpha particle, if very close to the daughter nucleus (which being inside certainly is) is strongly bound; but if it could "figure out" a way to get to a distance a little outside the surface of the daughter nucleus, the electrical repulsion would be bigger than the nuclear attraction and it would shoot out. The mechanism for getting away is called quantum tunneling and there is also a probability that an alpha particle will tunnel out and escape. Heavier nuclei (like Li) have a lower probability of formation and a lower probability of tunneling out if they do form. Roughly speaking, the probability of alpha decay is the product of the probabilities of formation and tunneling. You could also think of alpha decay as a very asymmetric fission of a heavy nucleus

QUESTION:
I am a high school physics teacher and am having some difficulty with part of the Compton Effect. I am trying to come to grips with whether the collision between the x-ray and the electron is elastic or inelastic? From what I have been able to find on the subject, when the x-ray collides with the 'whole' atom it results in an elastic collision and the x-ray leaves with the same frequency with which it came in with. On the other hand, when the x-ray collides with something closer to its own mass (an electron), it results in an inelastic collision and the x-ray is ejected with a lower frequency and energy. Any help you could provide would be greatly appreciated. I hate to think that I am not teaching it correctly and sending my students out into the world with misconceptions imparted to them by me.

ANSWER:
The Compton effect is elastic scattering of photons from some mass. Elastic does not mean that the energy of the incoming particle remains constant, it means that the sum of the energies of the incoming particle and the target remain constant. Assuming the target is at rest before the collision, after the collision it will recoil and carry away some of the energy which the photon brought in and the only place it can get this energy is from the photon. It is maybe easier to see this by thinking about classical particles. If a BB (photon) hits a bowling ball (whole atom), the bowling ball is almost at rest after being hit and therefore the BB has approximately the same energy (and speed) after the collision. If the BB (photon) hits a marble (electron), the marble will be moving after the collision so the BB must have lost energy (and speed). Both processes are elastic. An inelastic collision is one in which the total energy before and after are not equal.

QUESTION:
I�m having a problem determining the difference between Nuclear Fusion and Nuclear Fission. Listed below are some information that I�ve found. Could you clarify the major differences between the two, especially when it comes to matter that is heavier or lighter? Nuclear fusion - is the process by which multiple like-charged atomic nuclei (two or more joined together) join together to form a heavier nucleus. It is accompanied by the release or absorption of energy, which allows matter to enter into a plasma state. Nuclear fission - is a nuclear reaction in which the nucleus of an atom splits into smaller parts, often producing free neutrons and lighter nuclei, which may eventually produce photons (in the form of gamma rays).

ANSWER:
The important concept here is that if you find a reaction which results in less mass after the reaction than before the reaction, you will release energy because E=mc² and energy is conserved in an isolated reaction. Hence, if you lose mass, that energy must appear elsewhere. Where it usually appears is in kinetic energy, that is thermal energy of the reaction products. Iron is the most tightly bound nucleus, that is the mass of the average nucleon (a nucleon is a generic term for neutron or proton) is smallest. As shown in the figure, average nucleon mass steadily increases if you get heavier or lighter than iron. Hence, splitting a nucleus twice the size of iron or heavier results in mass loss and therefore energy gain; this is called fission. Similarly, fusing two nuclei half the size of iron or smaller results in mass loss and therefore energy gain; this is called fusion. Fusion is the source of energy for stars and hydrogen bombs. Fission is the source of energy for nuclear power plants and conventional nuclear bombs.

QUESTION:
During Nuclear Fission, how does the atom actually split? I understand that it becomes unstable with the absorption of a neutron but what I do not understand is how the atom can separate into two atoms. Eg. U-235 + n > U-236 > various products My main concern is how an atom can split.

ANSWER:
The atom does split, but the heart of the matter is the nucleus of the atom splitting. So, we usually do not call them atomic bombs anymore, we call them nuclear bombs (or, if you are George W. Bush, nuculer). Think of it like this: a nucleus is a very dense, very small object, often a sphere but sometimes more like a football, egg shaped. Uranium nuclei are more egg shaped. It also helps to think of the nucleus as a fluid. Now, if you add a neutron to ²³⁵U the resulting ²³⁶U nucleus is very excited. By this I mean that it is not just sitting there but vibrating wildly. So, think of this egg-shaped fluid vibrating so that it gets more elongated, in fact starts to develop a thinner "neck" in the middle, sort of like a peanut. What happens is that the two halves of this peanut-shaped thing break apart and, voil�, fission!

QUESTION:
Why does fusion stop at iron in a stars core? Is it temperature/pressure or something to do with the structure of iron?

ANSWER:
It is because iron is the most tightly bound nucleus and so any further fusion would require that energy be added. Heavier elements are made in supernova explosions. See an earlier answer for more detail.

QUESTION:
what evidence supports the contention that the strong nuclear interaction can dominate over the electrical interaction at short distances within the nucleus?

ANSWER:
The fact that nuclei stay together?

QUESTION:
I have read that cosmic ray particles can pass completely through the Earth. if this is true, is it possible to measure fluctuations in how consistently or rapidly they do so? The reason I ask is because, it seems that in the future we could produce a map of the Earth's substructure by reading such fluctuations like an x-ray machine. As the Earth rotated, I would imagine, it could be possible to triangulate on structures of varying densities, such as minerals, gas, or oil.

ANSWER:
The only type of particle which will pass completely through the earth is a neutrino and neutrinos are notoriously difficult to detect. They interact with matter very, very weakly (otherwise, how could they go through the whole earth). I can imagine no practical way they could be used like an "x-ray machine".

QUESTION:
My doubt is if protons repel protons how is it possible for them to be in the nucleus? if it is possible to separate a proton from a nucleus can we form new elements if done in large scale can we achieve creation of new elements?

ANSWER:
This is how we know that there is another force present besides the electrostatic repulsion. It is called the nuclear force or strong interaction. This force is very short ranged. That is, if the protons are not very close to each other, this force will will be very small and the repulsion will win out; but if they are very close, the nuclear force wins out and a nucleus may be held together. Neutrons also feel this force which is why neutrons are in nuclei.

QUESTION:
Is there any natural antimatter in the universe? I mean by "natural" antimatter that was not created by humans. Does it exist naturally in our universe, or somewhere else. Also, how exactly do we, humans, obtain antimatter? Do we get it from another universe? I know that matter cannot be created by the Law of Conservation of Matter, so how do we do it?

ANSWER:
Nuclei which have too many protons (or too few neutrons, depending on how you look at it) undergo β⁺ decay which turns a proton into a neutron, a neutrino, and a positron (an antielectron). This is a naturally occuring process. Antimatter can be created in an accelerator by colliding particles together; for example, smashing a proton into a nucleus can cause a proton-antiproton pair to be created. In fact, antiprotons are also naturally occuring in cosmic rays in the same way�an energetic proton from outer space enters the atmosphere and creates a proton-antiproton pair. There is no such thing as the "Law of Conservation of Matter"; matter can be created or destroyed.

QUESTION:
How did Chicago Pile 1 achieved a chain reaction? I know that they used purified graphite as a moderator and VERY pure uranium in the pile (reactor). The yellow cake was obtained from the Eldorado plant in Port Hope Ontario, which went through an ether process at the Mallinckrodt Chemical Works in St. Louis. Later some of the Mallinckrodt uranium was sent to the University of Iowa at Ames to be cast. Both the cast product and the Mallinckrodt product were used in the CP-1 matrix; the cast product, being purer, being placed closest to the center. During the testing, building up to the pile, they used a beryllium/radium neutron source, both in New York City and, later, in Chicago, to test the graphite as a moderator (as well as initiators for the atomic bombs). That I understand. However, when it came to the actual pile there is no mention of a beryllium/radium neutron source. It certainly appears that they relied on the uranium itself to initiate fission. But, how did they get the first neutron(s) to begin the chain reaction? Does U-235 undergo spontaneous fission? If so, it must be at a VERY slow rate and with a good moderator (graphite or heavy water). I've heard about spontaneous fission and the Flerov-Petrzhak discovery of sontaneous fission in 1940. Fermi must have known about this. So, did they use a radium/beryllium source or rely on spantaneous fission to start CP-1?

ANSWER:
It is indeed true that spontaneous fission is a rare event. On the other hand, there are one heck of a lot of atoms there and even very improbable events are quite possible at reasonable rates. Indeed, the first reactor, in Chicago, had no external neutron source but relied on spontaneous fission. Spontaneous fission can also be triggered by external radiation like cosmic rays. It took me a while to find a source which explicitly said this (see page 23).

QUESTION:
When 64Co27 (where 64 is the mass number of cobalt and 27 is the proton number) undergoes gamma decay it produces 64Co27 and energy which is in the form of the gamma ray. I don't get why the binding energy per nucleon has increased?, there has been no change in mass number?

ANSWER:
You have to conserve energy, right? The photon carries off an energy E, the original nucleus has energy M*c², and the final nucleus has energy Mc². Therefore, Mc²+E=M*c². This means that the final nucleus has less mass than the original nucleus, M<M*, which means that the final nucleus is more tightly bound, which means its binding energy per nucleon is greater. The binding energy per nucleon is determined by the structure of the nucleus, not the number of nucleons it has.

QUESTION: ;
Why do we always hear about critical mass in nuclear explosions. Shouldn't it really be critical density? It seems to me that the mass of fissionable material in a bomb is always there; it's when it gets compressed to some high density that the chain reaction occurs.

ANSWER:
Critical mass is one of those unfortunate misnomers (because it is not a mass) which has been passed down by common usage. (Another example would be electromotive force which is not a force.) Your suggestion that we call it critical density is not really very good either. I would call it critical geometry. For example, a hemisphere of uranium might not be a critical mass (that is, less than one neutron from a fission causes another nucleus to fission); but, if you put two such hemispheres together, they might be critical. When the material in a bomb is "compressed", it is brought into proximity with more material, it is not physically compressed to higher density.

QUESTION:
I'm a writer, working on a lyric essay about the space between things (kindof a lofty idea, I know... which is why it's a lyric essay). I've been trying to investigate what's in between the smallest particles, and I was hoping you could help me. I understand basic atomic structure, so you don't have to start from there. I'm wondering more (and forgive me if this is obnoxiously unscientific in its phrasing), what exists between an atom's nucleus and its surrounding electron cloud? Is it just emptiness? Am I riddled with little pockets of nothing? How can that be? Beyond that - what about the quanta? The quarks, up and down and so forth? Is there a space between these things (though, if I understand them properly... which I probably don't... they're energy, not physical structures)? I'm looking to investigate the scientific aspects of distance - large and small. Large, I get. It's just the small... the super tiny small. It's mind-boggling.

ANSWER:
Well, you are thinking too classically. At atomic scales, particles, like electrons and nuclei in atoms, are not discrete things which are localized. Rather, they are continuous distributions of what is referred to as "probability density", like a fog where the density of the fog is proportional to the probability of finding the object at that point in space. You are on the right track when you refer to an electron cloud, but the cloud is not like some shell, it is continuous like a fog and extends all the way in to the center of the nucleus; yes, the cloud is very diffuse inside the nucleus, but there is still a probability of finding an electron inside the nucleus. The probability density for the ground state of a hydrogen atom can be seen in the figure on the right (the relative size of the nucleus here is shown way too big). But, you can see that there is a region of higher density where we usually think of the "orbit" being but the electron is actually everywhere, just more likely to be in some places than others. In addition, we think of there being swarms of virtual photons. Virtual means that they appear and disappear very quickly; they may be thought of as the "messengers" of the electric force, called the quanta of the electromagnetic field. A similar thing happens inside the nucleus. There the density is much larger and, unlike the atom, pretty uniform, that is, the neutrons and protons (generically called nucleons) are almost equally likely to be found anywhere inside the approximately spherical volume and the density falls off very sharply at some radius so that the particles are much less likely to be found at, say, twice the average radius of the distribution than are the electrons in an atom. The quanta of the nuclear force are called mesons, and so these are popping into and out of existence inside the nuceus. But, unlike electrons, the nucleons and mesons are made of quarks, so the nucleus may be thought of as a smear of quarks and the quanta of their fields called gluons. So, you see, there is never anything which could be called empty space in matter. Even if you go into so-called empty space, what we might think was a vacuum containing no matter, there are constantly particles and their antiparticles popping into and out of existence, most commonly electron-positron pairs. This is called vacuum polarization and it is also going on inside the atom. Vacuum polarization has very minute but observable effects on the energies of atoms.

QUESTION:
In addition to studying cosmic rays, can we use detectors placed in space that utilize these same cosmic rays to create high energy collisions in order to search for fundimental particles in the same way the LHC is searching for these particles (ie: an LHC in space, without the need of the huge magnetic ring)?

ANSWER:
True, there are cosmic rays which have energies as high as achieved by the LHC, but to try to do serious particle physics with these particles is simply not practical. The two conditions imperative to perform useful experiments are high intensity beams of particles and control on the energy of the particles. You might get one energetic cosmic ray to hit your detector in a minute in space (I am really just guessing, but the point is that there are comparatively few) whereas you might get a billion per second in the LHC; because the events you seek to study happen with very low probability, you might have to wait a lifetime to get enough cosmic ray events to be statistically significant, probably longer. One of the most valuable ways to study particles and their properties is to see how their production depends on the energy of the collision; cosmic rays have a whole spectrum of different energies whereas the beam energy at the LHC is determined by how you set it.

QUESTION:
I am having a discussion with a colleague at the school where I teach. I believe that opposing beams are used in devices like the Large Hadron Collider because the collison of opposing beams (Beam A going clockwise and beam B going counterclockwise) have the additive energy of the the two beams, so the opposing beams give the apparent effect of approaching the speed of light without either particle having to be accelerated to such a great degree. My friend argues that if the two beams have the same energy it is no different than a single beam hitting a solid wall - beam A will be stopped dead in its tracks in either case - so there is no greater energy in either case. What is the real explanation?

ANSWER:
There are lots of misconceptions in what you both believe. Most importantly, experiments are not done by "hitting a solid wall", they are done by the beam hitting some other small particle, either some atom or nucleus; the most favored target is hydrogen which provides a proton target for the proton beam. This is the way particle physics used to be done, a beam of energetic protons hitting a target of stationary protons. But by making two proton beams collide, you gain much more than the added energy of the other beam (as I will explain below); if this were the case, it would be much cheaper to build an accelerator of twice the energy than to make a collider. And, the purpose of having two beams is not so that neither needs "to be accelerated to such a degree"; both beams are accelerated to as high as we can do it. The idea of many experiments is that we want to take the kinetic energy before the collision and have it available to create the masses of new particles. The classic example was the design of the first accelerator to make an antiproton (which has the same mass as a proton). If you collide an energetic proton with one at rest, you start with two protons and end up with three protons and one antiproton. Hence, you need to supply twice the rest mass energy of a proton for this reaction to happen, 2M_pc². So, you would just think you need to have the beam have an energy equal to 2M_pc², right? Well, it turns out that all the energy from a single beam is not available for making mass because, in addition to energy being conserved in the collision, linear momentum must also be conserved. Linear momentum is the mass times the velocity of the particle and, in the case of a single beam and a stationary target, there is a large linear momentum in the direction of the incoming beam; this same amount of linear momentum must be present after the collision. It turns out, if you conserve both energy and momentum, the incoming beam needs about 6M_pc²! About ²/₃ of the energy you put into the beam is "wasted" conserving momentum, just assuring that the system moves in the direction of the beam after the collision. But now look at the situation for a collider. Here, the linear momentum before the collision is zero because the beams move in opposite directions. Therefore all the energy of the beams is available to create particles. So a collider needs only M_pc² per beam to create a proton-antiproton pair.

QUESTION:
Inspite of the fact that is generally beleived that like charges repels each other while unlike charges.why is it possible to que all the protons in the nucleus of the atom together.

ANSWER:
A proton does not interact only with the electromagnetic force. It has mass and therefore interacts gravitationally with other masses. This, however, is negligibly small. It also interacts via the strong or nuclear interaction. This is a force felt by protons and neutrons, is much more attractive than the coulomb is repulsive at short distances. It is the strong interaction which holds nuclei together.

QUESTION:
Is it possible to find the exact a volume of a proton neutron or electron?

ANSWER:
Basically, no, because there is no well-defined surface. This is most clearly illustrated by looking at the density distribution of nuclei which are spheres similar but bigger than a proton (actually collections of protons and neutrons). Here we see the density of several nuclei plotted as a function of how far we are from their centers. Inside they are pretty constant, like a billiard ball, for example. But at the "surface" they are diffuse, not sharply dropping off like a billiard ball. (Incidentally, fm means 10^-15 m.)

QUESTION:
What is meant by "spin = 1/2" for elctrons? DOes it have anything to do with the spinning of electrons?What is the physical significance of the quantity "spin quantum number"?

ANSWER:
Elementary particles, like classical particles, may have angular momentum. A particle may have orbital angular momentum (like an electron orbiting around the nucleus in an atom or like the earth orbiting the sun) or it may have intrinsic angular momentum (like the earth spinning on its own axis or like the spin angular momentum of the elementary particle). When one goes to the microscopic level of elementary particles, angular momentum is quantized, that is only discrete amounts of angular momentum are allowed; for example, if the angular momentum quantum number of a particle is L, its angular momentum is [h/(2π)]√{L(L+1)} where h is Planck's constant. For a spin � particle, the intrinsic angular momentum quantum number is � and so the particles intrinsic angular momentum is h√3/(4π). Although it is sometimes described as spinning of the particle about its own axis, this is a classical picture which is useful only as a rough means of understanding what spin is. For example, it is impossible to do anything to change the spin of the particle.

QUESTION:
Can you please explain electron spin? It does not seem to fit in with the model I have been taught of a cloud of electrons 'orbiting' a central nucleus.

ANSWER:
The earth orbits around the sun. Therefore the earth has an angular momentum called orbital angular momentum. The earth rotates about its own axis. Therefore the earth has an additional angular momentum which we could call spin angular momuntum. The spin angular momentum has nothing to do with the orbital angular momentum.

An electron orbits around the nucleus (or, more sophiscatedly has a wave function which contains the information about the electron cloud). The electron has an angular momentum called orbital angular momentum, the information about which is also contained in its wave function. Like the earth, the electron has an additional intrinsic angular momentum which we call spin angular momuntum. It is as if the electron were spinning on its own axis (although that classical idea has problems if taken too literally). The spin angular momentum has nothing to do with the orbital angular momentum (or the electron cloud).

QUESTION:
Do electrons maintain a standard orbit about the nucleus?

ANSWER:
Actually, the idea of electrons being in well-defined orbits in an atom is just a pictorial way to qualitatively understand atomic structure. Originally Niels Bohr solved the puzzle of how atoms are constructed but his ideas later evolved into a much more complete theory of atomic structure. An atom consists of "clouds" of electrons around the nucleus, that is the electron does not maintain its identity as a point particle but becomes "smeared" over the volume in a way which is determined by the properties of the "orbital" it is in. This is quantum physics. However, if you say that the shape of the cloud represents the orbit, then, yes, electrons in one atom have the same distribution as in any other atom of the same element.

QUESTION:
I recently heard about the clever creation of Hydrogen 4.1 and its use to study reaction kinetics. However, I cannot find an explanation in any of the reporting, or original Science paper on why the muon orbits so close to the nucleus beyond "it's 200x heavier than electron." Surely they did not apply gravity to a Bohr model of a muonized atom, so why does the muon orbit closer?

ANSWER:
I actually never heard of hydrogen 4.1. Apparently what it is is a helium atom with one of the two electrons replaced by a negative muon. The radius of an orbit is inversely proportional to the mass in the Bohr model of the atom. It has nothing to do with gravity, it essentially enters from the mass in Newton's second law. So the muon, having about 200 times larger mass has a ground state orbit 200 times smaller. So, you may think of the "nucleus" as being of atomic mass of around 4 amu and charge of +1. Therefore this will look like a hydrogen atom with a much bigger mass.

QUESTION:
Fusion of (ionized) hydrogen molecules is done by increasing their temperature AND squeezing them using powerful electromagnets.(right?). If so, is it possible to "FUSE" them under normal room temperature just by indefinetly increasing the electro-magnetic force?. If so possible, what about "FUSION" under temperatures near 0 Kelvin ?

ANSWER:
The magnetic fields are not to "squeeze" them but to confine them. The high temperatures are required so that the positive ions have enough energy (that is enough speed) to overcome the electric repulsion from other positive ions. They need to get close enough to feel the nuclear force for fusion to occur and slow ions cannot do this. Furthermore, magnetic forces are perpendicular to the direction of motion so this force cannot squeeze; also, the magnetic force is proportional to the speed of the particle, so the slower the particle is moving (cold) the smaller any magnetic force is.

QUESTION:
Is it possible the reason why Electrons do not fall into the nucleus is; the closer an Electron gets to the nucleus it's charge is diminished, or it begins to become more positive and repelled? 2nd, the reason why it does not fly away from the nucleus is; the farther away it gets from the nucleus, it becomes more Negatively charged?

ANSWER:
Your suggestions are not possible. One of the most sacrosanct of all physical laws is conservation of electric charge. The charge of an electron never changes under any circumstances.

QUESTION:
Why is heavy water (D₂O) used in moderators of Nuclear Power Plants instead of normal water (H₂O)? If heavy water (D₂O) absorbs and slows down the neutrons emmited in fission what happens to it?

ANSWER:
The purpose of a moderator is to slow neutrons down so that they will be more probable to be absorbed by a fuel nucleus (e.g. uranium) and cause a new fission. And each new fission becomes the source of more neutrons, but they are fast and the moderator slows them down. If you use something like lead as a moderator, it would work poorly because the scattered neutrons would have almost no loss in speed (think of bb's bouncing off bowling balls). But, if you use something light, like hydrogen, then you are bouncing neutrons off something about the same mass which will quickly slow them down (think of a head-on collision between two billiard balls where the cue ball stops dead). So, hydrogen gas or liquid would be the best moderator, but hydrogen is very explosive, so we use something rich in hydrogen, water. But the problem is that a single proton can easily combine with a slow neutron (to create a deuteron) but that removes the neutron which we want to use to cause more fissions. The purpose of the moderator is not to remove neutrons, so we try the next lightest atom, deuterium, which is chemically identical to hydrogen but has a mass about twice as big. It is not quite such a good moderator, but it has a very small probability of absorbing neutrons. In the event that it does absorb a neutron, it becomes hydrogen with one proton and two neutrons and is called tritium. Tritium is radioactive and harmful to the environment.

QUESTION:
what keeps electrons energetically orbiting a nucleus?

ANSWER:
What keeps a satellite orbiting the earth? What keeps the moon going around the earth or the earth going around the sun? In all cases, once you give an object the energy required for a particular orbit, conservation of energy keeps it from changing. If the earth were to suddenly stop moving and drop into the sun, it would have far less energy; where did that energy go? To change the orbit you need to add or subtract energy to/from the object. What is particularly interesting about an electron in an atom is that an electric charge running around in a circle radiates energy (that is what an antenna is) and so the energy should radiate away and the electron fall into the nucleus. However, it does not and this observation started the whole branch of physics called quantum mechanics: one of the laws of nature is that bound objects are allowed to exist only in specific (quantized) energy states and therefore, if in such a state, a particle cannot radiate its energy away except all at once by going to a lower energy state. There is, however, a lowest state (called the ground state) and if the atom is in that state, it must stay there.

QUESTION:
If the half life period for say Uranium is constant then it should be that every atom of particular mass of Uranium degrade simultaneously and say after the half life for one atom whole number of atoms in that mass should get degraded and hence no more remians mass of Uranium but some thing else. Is it so?

ANSWER:
What half life means is that at that time half the number of original atoms will be converted into something else. It is useful only for a very large number of atoms because radioactive decay is a statistical process and you cannot know when any given nucleus will decay, only what the probability for decay is. After many half lives almost all the uranium will be gone.

QUESTION:
Do photons have mass? In one of your answers you say no. My penquin dictionary of science says, no mass. However, in the September 2006 issue of ASTRONOMY, the article, ASK ASTRO page 63. It is stated that the photon's mass is less than 10(-50)kilograms. Or 14 orders of magnitude smaller than a neutrino. Very low mass, yes, but there is mass. Is a photon massless?

ANSWER:
When we say that photons have no mass, we mean that every measurement we have ever made is consistent with their having no mass. However, it is often of interest to test the limits of our knowledge: how accurately do we know that the photon is massless? Most likely the article you were reading put an upper limit on what the mass of a photon could be based on experiments, 10^-50 kg.

QUESTION:
Could the comparison of an atom and a solar system exist with the current understanding of nano science. I.e. do electrons orbit the nucleus of an atom in relation with a nuclei's weight or mass? Do the nucleus' of atoms emit light?

ANSWER:
No. Gravity is not the force which holds the electrons in their orbit. It is the electrostatic force (the attraction between the negatively charged electrons and positively charged nucleus) which does it. Gravitational forces in an atom are entirely negligible compared to the electrostatic force. Mass, however, is not irrelevant, however. The mass of an electron is very small compared to the mass of the nucleus, so the nucleus, for all intents and purposes, remains fixed as the electrons orbit. If the mass of an electron and a proton in a hydrogen atom were equal, the two would orbit about a point halfway between them, much like a binary star.

Regarding your second question, yes a nucleus emits "light" but not light you could see. When an atom emits light it is often within the visible spectrum. The electromagnetic (called gamma rays) waves emitted from nuclei are much more energetic and thus of a far shorter wavelength than your eye can see.

QUESTION:
What is the angular momentum of the spin axis of an atomic nucleus? I have only heard about this in advanced physics. Is there any way one could reverse the spin axis of the angular momentum of an atom?

ANSWER:
The angular momentum of a nucleus is the sum of the angular momenta of its components. Each proton and each neutron has a spin (always 1/2 in appropriate units) and an angular momentum due to its orbital motion (always some integer in appropriate units); for each particle, its total angular momentum is the sum of these (always a half odd integer, i.e. 1/2, 3/2, 5/2...) If the nucleus has an even number of protons and of neutrons, the ground state angular momentum is always zero because it is energetically favorable for pairs to sum up to zero angular momentum. Only nuclei with odd numbers of particles or with an odd number of protons and neutrons have ground state angular momenta. The total angular momentum of a nucleus is usually referred to as its spin. To "reverse" the spin, you just need to flip the direction of the spin vector. So you should think of taking a spinning object and pointing its spin axis in the opposite direction, not stopping and reversing it; the effect is the same.

QUESTION:
When an alpha particle moves through an atom it will leave the atom undeflected if it is far enough away from the nucleus. What interactions are there with the electrons? Why do the electrons not attract and therefore deflect the positive alpha particle?

ANSWER:
Most certainly the alpha particle interacts with the electrons. But, the alpha particle has a mass about 8000 times bigger than the electron so it is like throwing bowling balls at bb's--the bowling ball is deflected only a miniscule amount. However, this is the mechanism by which alpha particles lose their energy going through matter; they have many collisions with electrons giving each electron a tiny amount of energy but eventually lose a large fraction (or all) of their energy. As you probably know, alpha particles don't go very far in matter--even a piece of paper can stop an alpha particle of a few thousand volts of energy.

QUESTION: ;
can you please explain nutrinos in basic terms?

ANSWER:
Many radioactive nuclei undergo a decay called beta decay. One kind of beta decay happens if a nucleus has too many neutrons (which have no electric charge). Somehow nature knows this and takes one neutron in the nucleus and turns it into a proton; but electric charge must be conserved, so the appearance of a positively charged proton necessitates the creation of a negatively charged particle, so an electron is ejected from the nucleus. It is this electron which is the "radioactivity". Back in the 1930s when beta decay was first discovered a most disturbing thing was observed: the electrons which were ejected came out with a broad range of energies. This was disturbing because each electron left behind the same thing (that is a nucleus with a particular amount of energy), so the range of electron energies implied that the energy was not conserved by beta decay. This principle, energy conservation, is so dearly held by us physicists that we invented a third particle, the neutrino, which had no mass and no electric charge and carried off the right amount of energy to insure that energy was conserved. A remarkable fact is that, because it interacts so incredibly weakly with matter, the neutrino was not experimentally observed until the late 1950s. So nearly 30 years elapsed before this hypothesized particle was observed, yet no self-respecting physicist doubted its existence!

Neutrinos have been in the news lately because of the so-called "solar neutrino problem". We believe that we understand very well what goes on inside the sun and yet the number of neutrinos which we observe on earth is considerably fewer than the number predicted by our models of stars. This problem has recently been solved by measurements which find that the neutrino is not really massless but has an extremely small mass (much smaller than the electron, the lightest of "everyday" particles). I will not attempt to explain how this solves the puzzle, but it does!

QUESTION:
Is there a way to change/slow down the rate of radioactive decay...for example, Tc99m half life is 6.01 hours. Is there a way to change that time, or slow it down?

ANSWER:
Not unless you slow down time itself, for example if the radioactive nuclei had a very high speed in the laboratory, they would live much longer because of time dilation.

QUESTION:
in my gcse physics lesson the teacher explained that an electron is given out by a nucleus when it changes during beta decay and a nuetron changes into a proton. Why is the electron made when the neutron changes? also is there such a thing as anti elements for example the exact opposite of hydrogen?

ANSWER:
There is also a third, chargeless particle, the neutrino, which is a product of neutron decay. It is not simple to answer a question like why the electron is ejected; that requires understanding the theory of beta decay (that is what this is called). One easy way to believe that an electron might come off is that the total electric charge of the universe must remain constant, that is you cannot create or destroy electric charge. If a proton appears, so does 1.6x10^-19 Coulombs of charge; to keep the total charge equal to zero, -1.6x10^-19 Coulombs of charge must appear which just happens to be the charge of an electron. And yes, antihydrogen (consisting of an antiproton and a positron) has been observed. Check out this story from CERN.

QUESTION:
I read that electrons at SLAC are accelerated to speeds of 0.999 999 999 948c. What are the specific products when electrons moving at this speed collide with other particles? Could you list some of the reactions produced at this speed? I would also like to know the applied EMF in volts needed for achieving the above speed, so that I can determine the energy of the accelerated electrons.

ANSWER:
The huge number of possible reactions depending on the target make it impossible to answer your first question. However, you second question can be answered pretty easily. However, you want to calculate the total energy E from the speed from which you can infer the voltage which must have been applied. The total energy is E=mc²/[1-v²/c²]^1/2. The arithmetic is a little tricky for v so close to c, but I find that E=4.9 GeV. This is the total energy, so to get the kinetic energy you must subtract the rest mass energy; but the rest mass of an electron is about 0.511 MeV=0.000511 GeV, so it is negligible. Hence, the EMF you seek is about 4.9 x 10⁹ volts.

QUESTION:
How can we measure the half life of radioactive meterial when the decay is completely random?

ANSWER:
The decay of any individual nucleus is indeed a random event, but it will occur with a probablilty. For example, one nucleus may have a 50% probability of decaying in the next hour whereas another may have a 50% probability of decaying in the next century. You learn this by watching a bunch of them and counting the number of decays you see in some time. Now if you have a very large number of radioactive nuclei (a typical macroscopic sample, say the head of a pin, would have something like 10²³ atoms in it), you can compare the rate of decays you see with the number of nuclei and thereby deduce the probability that any nucleus will decay in some particular time. The half life is the time it takes N nuclei to decay away to N/2 and, if you think about it, you can convince yourself that this is essentially a probability.

QUESTION:
I am trying to find out approximately what the speed of an electron around a hydrogen atom is? A teacher at my son's school said that an electron around an atom moves close to the speed of light? I found this difficult to swallow since according to my understanding of relativity, the mass of the electron would become very large at that speed and this would be the opposite of its known properties, ie small mass. I've done some research on the web and according to Bohr's theory, the speed of the electron should be approximately 1% that of light. But since his theory is not the prevailing theory and the Quantum Theory is, I was wondering if their is a probability distribution for the speed of electrons around an atom. It seems to me that if Quantum Theory can give us a probability of the location of electrons around a nucleus, then it can give us a probability of its speed around a nucleus.

ANSWER:
The teacher who claims the speed to be comparable to the speed of light is wrong; the Bohr theory gives the correct approximation of the electron speed to be 2.2x10⁶ m/s=0.7% c (c is the speed of light). However, as you note, the Bohr theory is not correct and quantum mechanics gives a better picture of what is actually going on. However, the essential details are the same--the most likely place the electron will be is near the Bohr orbit and the most likely speed is still around 1% c. Since the ground state of the H atom has a well defined energy, the sum of the potential and kinetic energies of the electron must always be the same. If the electron is at a smaller distance from the nucleus than the most likely, the potential energy will be smaller (more negative) so the kinetic energy will be larger (implying higher speed); similarly, if the electron is further away it will be moving more slowly. To give some perspective, if the electron were 1/100 of its most likely value (a very unlikely place to be found), its speed would still be less than 10% c. Again, using nonrelativistic quantum mechanics (Schrödinger equation) is not exactly right because of the neglect of the (small) relativistic effects, and the problem should be done relativistically.

QUESTION:
I am confused by some questions concerned with the difference among nuclear physics,particle physics,high energy physics etc.I'd love to study nuclear physics in UGA.But is it true that the particle physics and high energy physics are also related?

ANSWER:
Well, all of physics is ultimately one science. That is the beauty of physics--that the connections between all the various sub areas are pretty well understood (with a few rare exceptions). Nuclear physics, particle physics, and high energy physics are among the most closely related branches of physics.

Particle physics is the study of elementary particles, their properties, their constituents, and the theories which describe them.
High-energy physics is sort of ill-defined because there is no clear line where energy is high above it! Usually it refers to physics done at particle accelerators around the world which have the highest energies. In the 1950's 1 GeV (particles with energies equivalent to accelerating an electron across a potential difference of a billion volts) was high energy. Today it is more like a TeV (a thousand GeV) for high energy. Usually high energy physics is synonymous with particle physics but need not be. Sometimes high energy accelerators are used to study solid state physics or nuclear physics.
Nuclear physics was traditionally separate from particle physics, but if you think about it, that is kind of artificial, because what is a nucleus made up of anyway? From the 1930's through the 1980's the study of nuclear physics was just that--attempting to understand the structure of the nucleus. The two main means of achieving this are to look at the radioactive decay of nuclei and to look at particles scattered from nuclei (particles which have been accelerated in an accelerator).

By the 1980's and 1990's, great progress had been made in understanding nuclear structure in terms of a collection of protons and neutrons interacting via phenomenological forces. So much so that traditional nuclear physics is now thought by many to be an essentially solved problem. There are still lots of details and loose ends, and these can be interesting problems, but that is not the real essence of forefront science. Today, much of the field of nuclear physics has merged with particle physics/high energy physics. Now nuclear physicists are attempting to understand nuclear properties in terms of the underlying constituents of the protons and neutrons--quarks and gluons. In addition, inside the nucleus is being used as a place, a "laboratory" if you like, to do particle physics. Nuclear physics is still a vibrant and exciting branch of physics--it just isn't what it was 20 years ago and that is good!

QUESTION:
I understand that if negative and positive matter touch you get E=0, m=0, or in other words: Nullification. But what happens just before they touch? Since E=mc^2 can never be violated, and since everything in the universe ultimately comes from one field or another, when negative and positive matter 'almost' touch could local normal space/time be modified, creating a new E, m, and c? Relativity would still apply, but with a different c, which may be many times faster than our 'regular' c.

ANSWER:
When you refer to "negative and positive matter", I assume you mean "matter and antimatter". Positive and negative are words which usually refer to electric charge. "Nullification" is certainly not what usually happens when a positive and negative charge get together; for example, if you put an electron (negative) and proton (positive) together you get a hydrogen atom. A particle and its antiparticle usually have opposite electric charge (exceptions being neutral particles which have neutral antiparticles). When a particle and its antiparticle "touch", they annihilate, that is all their mass disappears (m=0). However, where you are incorrect is in saying that you get zero energy (E=0) because energy is conserved. E=mc² doesn't have to mean that if you have no mass then you have no energy. For example, radio waves carry energy but no mass. What normally happens when an electron and its antiparticle (a positron) annihilate is that two "photons", quanta of electromagnetic energy, are created which have all the energy which the electron and positron had before the annihilation, but none of the mass.

QUESTION:
Why does the neutron, a neutrally charged particle, have a magnetic moment? Isn't the magnetic moment defined as the direction of the magnetic field due to a rotational charge?

ANSWER:
Magnetic fields (which are how you detect magnetic moments) are caused by electric currents which are moving electric charges. The source of the field need not have a net electric charge to have a net electric current. The most obvious example of this statement is a simple current-carrying wire: although its net charge is zero, electrons move in the wire while the positively charged remainder of the wire does not; there is a current but no charge and there is a magnetic field. A neutron, as you know, has no net charge but experiments have shown that it has a charge density. Roughly speaking, the neutron has a positively charged core and a negatively charged surface. So if the whole neutron were thought of to be spinning as a rigid body (which is a naive but useful picture) there would be a magnetic moment opposite the angular momentum of the particle (in agreement with what is known about the neutron magnetic moment). This is because there would be more negative current than positive current since the negative charge is farther away from the spin axis than the positive charge so it moves faster.

QUESTION:
We know that during beta-decay, an neutron is converted to a proton and an electron. The electron was emitted leaving the proton inside the nucleus. So, it will increase the net charge of the atom by 1, changing it into an ion. As the decay goes on the net charge grows, the attraction between the ions is reduced due to the net charge produced. The radioactive material will explode eventually due to the repulsion mentioned above. However, it does not agree with our common experience that the material will not explode. How is it justified?

ANSWER:
This is more a question of scale and number than anything else. In order that the premise of your question be satisfied, let us assume that the radioactive source is in a vacuum and that it is thin enough that most of the electrons actually escape the source. A quite "hot" radioactive source would have one Curie (C) of radiation coming out and 1 C corresponds to 3.7x10¹⁰decays per second. That sounds like a lot, but consider how many atoms there are in the whole sample, something like 10²³, enormously larger than the number actually decaying over some time. So, on the average, the ions which form will be far separated from others. As time goes on, of course, the source will acquire a net positive electrical charge, so its electric potential relative to its surroundings will increase. Eventually, there will be a large enough potential difference that perhaps a spark will jump to neutralize the charge. Or, if the source is in the air, a large enough potential would lead to corona discharge to the surrounding air. In the real world it is pretty difficult to maintain a large static charge on something, and a static charge large enough to "explode" something just doesn't happen.

Oh, yes, one more issue: You have to consider what happens to the charge once it is created on one atom inside the volume of the source. If the material of the source is a conducting material, then any net charge must reside on the surface, so this puts the excess charge on the surface where it is easier for it to be neutralized by the environment.

QUESTION:
My question is, why do particles in a solid stick together? Also, what happens on the atomic scale to make the solid "break" apart when the substance is melted?

ANSWER:
Think of an atom as a tiny positively charged nucleus surrounded by a "cloud" of electrons. When these two atoms approach each other there is, when they are farther apart than their sizes, no force because each is electrically neutral. As the clouds begin to overlap, there is a repulsion due to the force between electrons, but there is also an attraction because the electrons in one atom are attracted to the nucleus of the other. In most cases there is a region, that is a range of separations, for which the attractive force is larger than the repulsive force and the result is that the two atoms may bind together. This is what causes molecules to form and what causes solids to form. However, this sticking together depends upon the kinetic energy of the individual atoms; and, the average kinetic energy of the atoms is simply a measure of the temperature of the collection of atoms. For example, consider a hydrogen molecule H₂: think of the binding as a tiny spring connecting the two atoms and, when the H₂ gas is heated up, some of the kinetic energy is taken by the oscillating pairs of atoms such that the amplitude of oscillation may be large enough to "break" the spring. This same thing happens in a solid where we may think of all the atoms as being rigidly bound to their neighbors. As the solid is heated up, the bonds break and the solid is no longer a solid. At first it is a liquid where the atoms or molecules move around in close proximity to their neighbors with bonds being constantly in a flux of bonding and breaking again. As more energy is added to the liquid, individual atoms or molecules acquire enough kinetic energy to escape entirely forming a gas.