Coulomb barrier and proton evaporation


by cesiumfrog
Tags: barrier, coulomb, evaporation, proton
cesiumfrog
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#1
Dec10-12, 09:19 PM
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Why is it that neutrons evaporate from nuclei more easily than protons do?

Intuitively, since protons are electrostatically repelled (in addition to whatever nuclear forces they have in common with neutrons), one would expect protons to be ejected more readily than neutrons. (Maybe this is even what does happen for small nuclei, but apparently not large nuclei.)

It seems to be said in common parlance that the Coulomb force/barrier acts to contain the protons. Which is counter-intuitive.

On the other hand, at least some textbooks acknowledge the coulomb force trying to push protons away from others in the nucleus, and so they infer that the nuclear force (e.g., residual strong force) must act more strongly on protons than what it does on neutrons. (At least in big nuclei, "beta stable nuclei", nuclei with an excess of neutrons..) So how does this nuclear force distinguish between protons and neutrons?
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tom.stoer
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#2
Dec11-12, 12:38 AM
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I don't see in which process neutrons evaporate from nuclei
K^2
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#3
Dec11-12, 04:52 AM
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Quote Quote by tom.stoer View Post
I don't see in which process neutrons evaporate from nuclei
Neutron Emission

This does happen with protons as well, but yes, it is far more likely to happen to a neutron in a heavy nucleus, and the reason for it is higher potential barrier for the protons. Yes, it's counter-intuitive, but actually pretty straight forward. Consider nucleon potential of this form. (I made up the shape, but it's qualitatively correct.)



For neutrons, this is all there is. For protons there is an additional repulsive potential. Lets just throw in a 1/r from center. It's not quite right at the center, but we are only interested what happens near the edges of the nucleus.



Certainly, the energy difference between the ground state and barrier for protons has only decreased. But we aren't interested in a proton or neutron in the ground state of this potential. We are interested in protons and neutrons with the highest energy in this nucleus.

Why is it that heavy nucleons have much higher number of neutrons than protons? Because a nucleus with same number of both would have protons with much higher energy. That results in protons undergoing beta decay and becoming neutrons. If there are any protons with significantly higher energy than neutrons, there will be beta decay.

That's where things start getting interesting. The energy of the most energetic proton and neutron are going to be roughly the same. The potential energy for protons, however, has increased everywhere. That means that relatively to the highest point of the potential barrier, your most energetic proton is sitting lower than neutron is sitting relative to its barrier.

In other words, thanks to electrostatic repulsion, the protons are fewer, but because they must have the same total energy as neutrons, they have a higher barrier to overcome in order to escape thanks to their repulsion.
Attached Thumbnails
Potential1.png   Potential2.png  

andrien
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#4
Dec11-12, 06:52 AM
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Coulomb barrier and proton evaporation


@ k^2,will it be right to call it evaporation type thing?
K^2
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#5
Dec11-12, 01:31 PM
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I don't see why not. So long as you are happy with both thermal excitation and tunneling contributing to particles escaping a bound state as part of your definition of evaporation. Otherwise, the physics is very similar to classical evaporation. The net increase in potential energy of the system is covered by the decrease in temperature of the nucleus and increase in overall entropy.

I'm not sure I've ever actually heard anyone call it evaporation before, but I don't work with nuclei, so a lot of their jargon is foreign to me. I've heard term "evaporation" used for just about every other kind of particle escaping bound states, however, so I don't see why it wouldn't apply.
Bill_K
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#6
Dec11-12, 01:47 PM
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Thanks
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Yes it is, primarily when the nucleus is so far off the stability line that several neutrons are emitted, such as from a fission fragment.
cesiumfrog
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#7
Dec14-12, 11:21 PM
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Thanks for the replies!

So are Blatt & Weisskopf mistaken when, comparing protons to neutrons, they write "The protons are more strongly bound by the nuclear forces in order to compensate for the electric repulsion energy"? http://books.google.com.au/books?id=...ongly+bound%22

In classical physics if I had two kinds of equal-mass particles all bouncing around together in the same gravitational potential-well (e.g., a bucket), and one kind had homogeneous electric charge (or some other simple mechanism forcing them toward the outside) then I would intuitively expect those repelled particles to be the most likely kind to actually get ejected. Does everyone agree with me, or need I run a simulation to check?

K^2, in your diagram (by your choice of where to draw the zero levels) you seem to be comparing protons and neutrons on the basis of their kinetic energy after they escape to infinity. To me, it makes more sense to compare them on the basis of their kinetic energy at some given radius within the well. Is it not the case during billiard ball collisions that their kinetic energies on average get distributed evenly, without distinction as to what their respective potential energies are (i.e., never minding whether some have springs attached nor electric potential etc)? Or to put it in other words:
Quote Quote by K^2
The [kinetic] energy of the most energetic proton and neutron are going to be roughly the same. The potential [well] for protons, however, has [become shallower]. That means that relatively to the highest point of the potential barrier, your most energetic proton is sitting [higher] than neutron is sitting relative to its barrier.
K^2
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#8
Dec14-12, 11:31 PM
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So are Blatt & Weisskopf mistaken when, comparing protons to neutrons, they write "The protons are more strongly bound by the nuclear forces in order to compensate for the electric repulsion energy"?
No, that's exactly the same thing. Kinetic + Nuclear + Coulomb potential energy of protons = Kinetic + Nuclear + Coulomb potential energy of neutrons. Because Nuclear PE is the same for both, and Coulomb PE is zero for neutrons, the KE of protons is lower. That means, they sit lower in Nuclear potential. That means, they are more tightly bound.

Just two sides of the same coin.

Quote Quote by cesiumfrog
K^2, in your diagram (by your choice of where to draw the zero levels) you seem to be comparing protons and neutrons on the basis of their kinetic energy after they escape to infinity. To me, it makes more sense to compare them on the basis of their kinetic energy at some given radius within the well.
Yes, I chose infinity, because it's the least messy choice. You can choose any point you like, but regardless of choice, the neutron and proton potential energies must be different by the coulomb term for the later. Remember, kinetic energy does not depend on choice of zero potential. So this difference must be consistent for your description to be consistent. Since the difference is zero at infinity, it's the natural choice, but you can work it out at an arbitrary point.
cesiumfrog
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#9
Dec15-12, 01:22 AM
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K^2, though I respect that this is supposedly much closer to your area of speciality than mine, I'm really having difficulty making sense of your reasoning (in context). For example, I don't see why the total energies would stay the same, nor why the difference between electrostatic potentials must necessarily be chosen to be zero at infinity.

What did you conclude in my toy classical example? Or say we had a cup containing a mix of iron and copper marbles of the same masses, with a very strong magnet afixed atop the cup handle, and strap this to a bicycle speeding down an unsealed road: which kind of marble would you predict more to get ejected from inside the cup?

Looking to another textbook, Cottingham & Greenwood notes "For nuclei with more neutrons than protons the contribution of the strong nucleon-nucleon interaction to the potential is more attractive for the protons than for the neutrons, since a proton in such a nucleus is, on average, more subject to the neutron-proton interaction". If you still think what you said was valid, how would you accomodate Blatt & Weisskopf's statement that for light nuclei with equal numbers of protons and neutrons "it is actually easier for a proton to emerge than for a neutron"?
K^2
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#10
Dec15-12, 02:19 AM
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The energies must be the same due to beta decays. In your toy problem, iron marbles can't turn into stone marbles to lower their energy. The nucleons can and do. If you have exactly the same number of protons and neutrons in a heavy atom, the protons will start to decay to neutrons to reduce the overall energy. And you eventually end up with excess of neutrons, which are energetically favored. So in a stable nucleus, the energy of protons and neutrons is roughly the same.
cesiumfrog
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#11
Dec15-12, 06:02 PM
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So let the iron marbles be able to turn to stone! Wouldn't you still predict the remaining portion of iron marbles to be the ones with the best chances of coming out of the cup? I can see how the weak nuclear force can directly determine whether or not a nucleus is stable to beta decay, but assuming a particular nucleus doesn't beta decay first, I don't see any way the weak force would directly affect which kind of nucleon is more likely to be emitted.

To emphasise my point of conflict here, it is that you have shown two plots in which the residual-strong-nuclear component of the potential is identical for both neutrons and protons (i.e., only the coulomb potential differs), and used this to assert that neutrons are always more likely to be ejected. I think what those plots actually show is that in the case of Z=N (as tends to occur stably for calcium and lighter elements) the protons are more likely to be ejected. But for neutrons to be more likely to be ejected, we would need a different pair of plots in which the residual-strong-nuclear component of the potential is stronger for protons than it is for neutrons (as in fact is the case in nuclei with N>>Z due to the inequality of residual-strong forces between same-same and otherwise pairs of nucleons).

I can follow B&W's argument that either a proton or a neutron extracted from a large beta-stable nucleus would have similar amounts of kinetic energy as they approach infinity (because the difference in binding-mass-energies liberated from this nucleus equates with what is liberated in beta decay of a nucleus very similar to our original nucleus, which must be a small amount of energy because of the closeness to beta-stability). I don't grok your reason for stating that the difference between the total energy of a nucleon inside the nucleus and the potential energy of the same nucleon at infinity should be equal for protons to whatever it is for neutrons. I can see a good reason for the protons and neutrons inside the nucleus to have similar amounts of kinetic energy (equipartition theorem of thermodynamics), although obviously your statements stand in contradiction with this.
K^2
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#12
Dec15-12, 07:15 PM
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Quote Quote by cesiumfrog View Post
So let the iron marbles be able to turn to stone! Wouldn't you still predict the remaining portion of iron marbles to be the ones with the best chances of coming out of the cup?
No add the fact that steel marble and stone marble can occupy the same spot in the cup. And also, the fact that strength of the magnet changes with number of iron marbles. Finally, the magnet is pushing steel marbles out (you can achieve that with alternating current) rather than pulling them. You'll end up with steel marbles ending up fewer, closer to the bottom, and less likely to escape due to vibration.
cesiumfrog
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#13
Dec17-12, 04:41 PM
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Perhaps may I ask slightly different questions:
  1. Why do nuclei with equal ratios of protons and neutrons (e.g., in elements lighter than calcium) tend to emit protons more easily than neutrons?
  2. Why is the proton-neutron force stronger than the proton-proton and neutron-neutron residual-strong-forces?
  3. What conditions are necessary to violate the equipartition theorem of statistical mechanics (i.e., why should protons have much less kinetic energy than the neutrons in the same nucleus)?
K^2
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#14
Dec17-12, 10:57 PM
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Good questions. Let me start with the last one.

The kinetic energy of nucleons in nucleus have almost nothing to do with equipartition theorem. These aren't thermal excitation energies. The energies there are keV, if not MeV ranges. Compare it to thermal energies in fractions of eV. Kinetic energy of protons and neutrons is due to shell structure. Basically, think of Hydrogen atom as your analogy. Lower levels are occupied. Higher levels have higher kinetic and potential energies.

Despite the coulomb repulsion, the shape of potential for neutrons and protons within the nucleus is roughly the same. That means the shell structure will be similar as well. That means, if there are more neutrons, they occupy higher levels, and they have way more energy.

Equipartition will apply to thermal smearing of occupations of higher levels, and there, yes, the distribution in energy relative to ground state will be equal for protons and neutrons.

With this in mind, answer to first question is also pretty clear. If you have same number of protons and neutrons in a nucleus, their kinetic energy is the same. The nuclear potential energy is also the same. So the difference will be made by coulomb repulsion. The barrier will be slightly lower for protons.

Finally, we go to your second question. The answer is that it isn't. Not exactly. The nuclear force itself is the same. However, p-p and n-n will experience Pauli repulsion, while p-n will not. That results in p-n being more tightly bound in a nucleus.
cesiumfrog
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#15
Dec21-12, 01:00 AM
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Ah.

So some authors combine residual strong force with Pauli degeneracy pressure, expressing this as a difference between the strength of proton-neutron and same-same interactions, which makes shallower the potential well of whichever kind of nucleon is in excess. Others prefer just to say that an excess of neutrons requires the neutron-states to be occupied higher than the proton-states (when the nucleus is in its ground state). Either way, an excess of one kind of nucleon decreases the relative gap from the least-bound nucleon of that kind to the top of its well.

The Coulomb interaction does unambiguously make it easier for protons to escape the nucleus (as is intuitive; the only time it presents directly as a barrier is to scattering but never to emission or evaporation). But this can always be outweighed by a sufficient excess of neutrons.

The weak force has absolutely no bearing on which kind of nucleon would be more likely to get emitted from a given nucleus. But it can tell us something about the kinds of nucleus which would be stable enough to have come under our consideration. And this is where comparison with neutron and proton potentials at infinity becomes relevant. If removing one nucleon while adding another of the other type could have liberated much energy to radiate away, then that tranformation likely would have happened already, so we can expect to encounter only nuclei where the least-bound of each kind of nuclei has a roughly similar total energy.

So in a stable heavy nucleus the neutrons have far more kinetic energy, but can't exchange it to the protons because the lower neutron states are already taken. If something like a cosmic ray excites a proton then the energy is likely to dissipate (evaporating neutrons, beta-decaying, re-radiating) before any proton tunnels past the strong force. (The Coulomb force assists the proton against the strong force, but it also provides incentive to turn into a neutron instead.)


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