# Neutron or proton emission?

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## Main Question or Discussion Point

Hello,
For a given excitation energy, why a nucleus "choose" to evaporate preferentially a neutron or a proton?
I mean, let us take for example the isotope 208Pb. Its neutron separation energy is 7367.87 keV while its proton separation energy is 8004 keV. If this isotope has an excitation energy of 9 MeV, it has enough energy to evaporate one proton or one neutron but I am pretty sure that it will much more often a neutron instead of a proton. Why?

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DrSteve
Gold Member
Hello,
For a given excitation energy, why a nucleus "choose" to evaporate preferentially a neutron or a proton?
I mean, let us take for example the isotope 208Pb. Its neutron separation energy is 7367.87 keV while its proton separation energy is 8004 keV. If this isotope has an excitation energy of 9 MeV, it has enough energy to evaporate one proton or one neutron but I am pretty sure that it will much more often a neutron instead of a proton. Why?
For a nucleus near its ground state the answer is the proton has to tunnel through the Coulomb barrier. See http://physics.stackexchange.com/questions/56094/coulomb-barrier-and-proton-evaporation for a more detailed discussion.

berkeman
Mentor
The Mentors would like to re-word the OP statement for clarity (and we've edited the thread title for clarity):
For a given excitation energy, why a nucleus "choose" to EMIT preferentially a neutron or a proton?
I mean, let us take for example the isotope 208Pb. Its neutron separation energy is 7367.87 keV while its proton separation energy is 8004 keV. If this isotope has an excitation energy of 9 MeV, it has enough energy to EMIT one proton or one neutron but I am pretty sure that it will much more often EMIT a neutron instead of a proton. Why?

For a nucleus near its ground state the answer is the proton has to tunnel through the Coulomb barrier. See http://physics.stackexchange.com/questions/56094/coulomb-barrier-and-proton-evaporation for a more detailed discussion.
I agree for a nucleus with an excitation energy below the proton separation energy. But my question is about if the nucleus has an excitation energy a bit higher than this separation energy (as I explained with 208Pb nucleus)

@berkeman: thanks for the correction, you were right.

mfb
Mentor
"Proton separation energy" still means you have to tunnel through the Coulomb barrier. The energy is sufficient to do so (that's the overall energy condition), but the process is suppressed. It is exactly the same situation as alpha decays: if a decay is possible it can still have a very long lifetime.

"Proton separation energy" still means you have to tunnel through the Coulomb barrier. The energy is sufficient to do so (that's the overall energy condition), but the process is suppressed. It is exactly the same situation as alpha decays: if a decay is possible it can still have a very long lifetime.
No, "proton separation energy" means that the nucleus has enough energy to emit spontaneously a proton. Taking into account the Coulomb barrier explains why the proton separation energy is higher than the neutron one. You can see for example the figure 3, page 161 from this pdf (for the nucleus within this picture if the excitation energy is let us equals to 15 MeV, then there is no more barrier). So the question remains: if a nucleus has enough energy to emit spontaneously either a proton or a neutron, why does it "choose" to emit preferentially a neutron.

mfb
Mentor
Proton emission becomes possible as soon as tunneling through the barrier releases energy.

It becomes very fast if this energy is larger than the Coulomb barrier.
Ep in figure 3 is the 8 MeV value. I don't know how large the Coulomb barrier is for lead, but it is non-zero - even if it is smaller than 8 MeV, your neutron still has a larger phase-space.

Edit: fixed formatting

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Proton emission becomes possible as soon as tunneling through the barrier releases energy.

It becomes very fast if this energy is larger than the Coulomb barrier.
Ep[/sup] in figure 3 is the 8 MeV value. I don't know how large the Coulomb barrier is for lead, but it is non-zero - even if it is smaller than 8 MeV, your neutron still has a larger phase-space.
My question is about exactly that point. Above a given energy, let us say 50 MeV just to be sure to be much higher than the Coulomb barrier. Then, with such energy, I am pretty sure that the nucleus will emit a neutron rather than a proton (for 208Pb I am not absolutely sure but for 250Cf, I am certain). Why?

DrSteve
Gold Member
(for 208Pb I am not absolutely sure but for 250Cf, I am certain). Why?
With 50 MeV of excitation energy the probability of neutron vs proton emission for 208Pb should be nearly identical. The lower the excitation energy the more the separation energy difference comes into play - by the time you're down to 9 MeV of excitation the .6 MeV difference completely dictates the outcome because of the tail of the Coulomb barrier (study the formula on p 161, which applies only to protons).

mfb
No, I completely disagree, proton emission is always (often?) less favorable compared to neutron emission even at higher energy. For example let us consider now the spallation reaction as what it is done at GSi, Germany. They accelerate 208Pb at let us 1 Gev/nucleon. Lead beam impinges on target and nucleons are removed from the lead. Lots of neutrons are evaporated (tens) while only few protons are evaporated. That is why we can produce nuclei until proton drip-line and even further but we are not able to produce 201Re (although this nucleus has "only" 7 protons less than 208Pb). Cross section decrease extremely quickly when we want to remove one neutron compared to the removal of one neutron. I guess it is the same phenomenom in both cases.

The formula p161 does not matter at all here. It describes the tunneling probability since it is matter of proton radioactivity which means poroton emission from the ground state or at least from a state below the proton separation energy. What am I talking about happens above the Coulomb barrier.

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DrSteve
Gold Member
No, I completely disagree, proton emission is always (often?) less favorable compared to neutron emission even at higher energy.
Yes, because of the Coulomb barrier. See e.g. Bromley's Treatise on Heavy-Ion Science: Volume 8: Nuclei Far From Stability p 108. I'm a nuclear physicist and did my dissertation on transfer reactions in deformed nuclei- perhaps we could discuss this further off line, if needed.

Hmmm, I still do not understand why, at high energy, Coulom barrier still matter. If the Coulomb barrier is let us say equal to 20 MeV and the excitation energy is 100 MeV, then Coulomb barrier effect should become negligible. Let me think about that.

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So, I did not find yet the full explanation but just to illustrate what I meant above, let us look at this picture

In this example, both neutron and proton potential wells are shown for the case of 116Sn. We see clearly that the proton potential is higher due to the Coulomb force; for 116Sn, the Coulomb energy is about 10 MeV. As tin has 50 protons, the last shell filled by protons is about at -10 MeV. About neutrons, neutrons in outer shell are around -9 MeV. So, if 116Sn has an excitation energy of let us say 200 MeV, it becomes possible to evaporate both proton or neutron. Let us say both neutron and proton separation are equal to 10 MeV. The nucleus can then evaporate 20 nucleons. For the last two nucleons, I agree that the Coulomb barrier matters since there is not enough energy anymore to "pass" above the Coulomb barrier and so neutron evaporation will be favored. Yet, for the 18 others, how the nucleus choose to evaporate proton or neutron (or deuteron, alpha, gamma and so on)?

Currently, I am investigating the Heiser-Feshbach statistical model since the answer is likely highly related to the decay channel width. So if you have some clue about this model, let me know :)

DrSteve
Gold Member
I admire that you are still pursuing this topic. I learned much of my nuclear physics from Nilsson's papers.

I believe the answer is manifest from the two potentials. The neutrons need ~ 9 MeV to be emitted by the compound nucleus; protons around ~ 21 MeV, since they otherwise have to tunnel through the Coloumb barrier so explicitly shown above. The probability of any one nucleon receiving 21 MeV when only 200 MeV of excitation is involved is very small; the probability of any one nucleon receiving 9 MeV is much, much larger. Unfortunately, that's not enough energy for the protons to escape the potential.

mfb
Mentor
The probability of any one nucleon receiving 21 MeV when only 200 MeV of excitation is involved is very small
I think a key question then becomes "is the energy distributed (or even in thermal equilibrium)"? If yes, then the effect is clear - with an average of 2 MeV many neutrons will reach 9 MeV while few protons will reach 20 MeV. If it is non-thermal, things are less clear.

DrSteve
Gold Member
I think a key question then becomes "is the energy distributed (or even in thermal equilibrium)"? If yes, then the effect is clear - with an average of 2 MeV many neutrons will reach 9 MeV while few protons will reach 20 MeV. If it is non-thermal, things are less clear.
The compound nucleus is, by definition, in complete thermal equilibrium. See e.g., http://www.epjconferences.org/articles/epjconf/pdf/2014/06/epjconf_cnr2013_00012.pdf [Broken]

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mfb
Mentor
That link seems to be broken.

How can the nucleus follow an external definition?

DrSteve
Gold Member
That link seems to be broken.

How can the nucleus follow an external definition?
Perhaps poor choice of wording. See the attached file.

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mfb
Mentor
Well, that article discusses hot nuclei only. Do we know the mentioned collision process lead to those? Apparently, but that is a result of observations - you cannot define it to be like that.

DrSteve
Gold Member
Well, that article discusses hot nuclei only. Do we know the mentioned collision process lead to those? Apparently, but that is a result of observations - you cannot define it to be like that.
There are three time domains, if you will, for nuclear reactions, that differ by how much time the beam nucleus spends in the vicinity of the target nucleus. The shortest time domain is for so-called knockout reactions, where the incoming nucleon directly interacts with one of the nucleons in the target. This timescale is ~ the time it takes the beam particle to traverse the nucleus ~ 10-21s. The next timescale applies to transfer reactions, where the beam nucleus may make multiple orbits around the target nucleus and there is sizeable energy sharing. Nucleons can be transferred into out out of the beam nucleus depending on the N/Z ratio of each. There is a variant of this called deep inelastic reactions, where the timescale is even longer and there is quasi-equilibrium with neutron evaporation. This topic was the subject of my PhD thesis. The longest timescale ~ 10-16s is the formation of the compound nucleus, where the excitation energy of the fused nucleus is very high and complete equilibration is obtained. The heavier the beam and target pair the more likely that compound nucleus formation will occur (though, perversely, a single neutron can incite in a n + 235U fission reaction).

So, as you allude to, the details of the reaction are crucial. I'm taking an educated guess that a 20 MeV proton or neutron incident on 116Sn would have roughly equal chance of knocking out a proton or neutron if the Q-values were similar. By contrast, a 200 Mev 48Ca nucleus incident on 208Pb is going to create a compound nucleus with the evaporation of a handful of neutrons but no protons.

mfb