B What is the equation for proton emission after muon capture in heavy nuclei?

[aisha hamzah]

Can anyone tell me what is proton emission equation after muon capture for heavy nuclei?is it same with neutron emission equation??

 

[mfb]

What do you mean by "proton emission equation" or "neutron emission equation"? The energy of the emitted particle? That will depend on the nuclide.
Do you mean muon capture in the same way as electron capture?

 

[aisha hamzah]

After negative muon is capture, some of muon decays to electron and emit neutrino. Then, neutrino take of most of the energy and leaving 10-15% for neutron to recoil process following the reaction. The nucleus and neutron might have energy distribution, E(Q) which depend on momentum distribution of the capturing proton. Then, the probability of proton emission is small compared to neutron emission due to effect of coulomb barrier. So, in this case, I want to calculate the proton emission after muon capture process.

 

[mfb]

Okay, so I guess you mean the equivalent of electron capture - the process in the nucleus. There is no electron involved in this process.

Then, the probability of proton emission is small compared to neutron emission due to effect of coulomb barrier.

And due to the fact that no proton is produced. You would need an energy transfer from the new neutron to the proton. Not impossible, but I would expect that to be very unlikely, and very difficult to calculate theoretically.

Where do you need this?

 

[snorkack]

Okay, so I guess you mean the equivalent of electron capture - the process in the nucleus. There is no electron involved in this process.And due to the fact that no proton is produced. You would need an energy transfer from the new neutron to the proton. Not impossible, but I would expect that to be very unlikely, and very difficult to calculate theoretically.

I, on the other hand, would expect it to be quite likely.
The initial reaction is
p+μ→n+ν_μ
Since the energy released is over 100 MeV, most of it goes to neutrino. But something like 15...20 MeV should go to the recoil on neutron. Which exceeds the binding energy of the neutron.
What next?
The original neutron might fly straight away and leave the rump nucleus in ground state.
But I would not expect energy transfer by strong interaction to be "very unlikely". After all, the neutron might form inside nucleus or, if forming on surface, start out in the direction through the nucleus.
So what next?
Energy transferred to one other nucleon?
Or energy transferred to many other nucleons? 15 MeV is enough to unbind one nucleon, but divide it in a nucleus of 150 nucleons and each nucleon gets just 100 keV - much less than its binding energy. Will the nucleus then emit nucleons, or photons?

There are similar common processes. Like. the common fusion reaction
d+t→α+n
gives neutrons at 14 MeV
So if a heavy nucleus meets a 14 MeV neutron, there are many options:
n,γ
n,nγ
n,p
n,2n
n,np
n,f
etc.
But how do you derive from first principles the branching ratios between the above processes?

 

[mfb]

Expecting something doesn't make it true.

https://journals.aps.org/pr/pdf/10.1103/PhysRev.124.1602

In light nuclei, the created neutron carrying this energy usually leaves the nucleus without further interaction. In intermediate and heavy nuclei the neutron may divide this energy between the other nucleons and a compound nucleus is formed. The excitation energy is then lost by evaporation of neutrons and to a lesser extent by emission of charged particles.

Nuclear capture of negative μ mesons in photographic emulsions

About 24 000 meson tracks which stopped in the emulsion have been studied. In 591 cases (2.4 percent) the stopped meson is accompanied by the emission of one or more charged particles.

Mainly silver and bromine as target. The number includes ~20% alpha particles and ~10% background.
Figure 4 has the spectrum, most protons have 10 to 15 MeV.Muon capture in nuclei, L. L. FoldyJ. D. Walecka
They don't even consider proton emission because it is so rare.

Muon capture rates in complex nuclei, M.Eckhause, R.T.Siegel, R.E.Welsh, T.A.Filippas
They veto charged particles because they need the neutrons for a reasonable signal/background ratio.