Yes - if the proton can escape - more likely an alpha particle - it gets additional energy
due to the electrostatic repulsion of the alpha and the remaining nucleus.
There's more to it than just the energy. For example, let's take a target nucleus that
has Z protons and N neutrons. A = Z + N. Let's assume Z is odd and N is even.
Then A will be odd, and the [tex]\delta[/tex] in the expression above is zero. The nucleus will
have a binding energy given by the above.
Now the nucleus captures a neutron. We now have to compute a new binding energy
for the compound nucleus. In this case, the value of A --> A + 1; so the new A is even.
The number of neutrons N --> N+1; which is now odd. When A is even; but both Z and
N are odd; the value of [tex]\delta[/tex] is negative. [The [tex]\delta[/tex] term is called the "pairing term"; and is
due to a quantum mechanical effect.]
So in effect, the compound nucleus "looses" some binding energy. [ This doesn't mean
that any real energy is being destroyed; it just means that the ground state energy for the
compound nucleus is higher than it would have been otherwise.]
So you now have a nucleus that not only has a bunch of energy that the new neutron
brought in - but also its ground state energy has changed. These two effects coupled
together may mean that the new compound nucleus is unstable. The nucleus has to
now find a stable state.
Most likely, a nucleus that absorbs a neutron will be unstable with respect to [tex]\beta-[/tex] decay.
That extra neutron will turn into a proton, an electron, and an anti-neutrino; and the latter
two will be ejected and the nuclide transmutes to one with the next higher atomic number
Z+1 due to the new proton.
Note that the daughter nuclide with atomic number Z+1 will have more electrostatic repulsion
than the parent which had atomic number Z. There will be MORE Coulomb "unbinding energy".
However, the nuclear effects are more important than the Coulomb effect. Since the original
value of Z was odd; the new value Z+1 will be even. The neutron number N will go back to
its original value which is even. So the daughter nucleus will have an even number of protons,
and an even number of neutrons. This is more important for stability of the nucleus than is
the extra Coulomb repulsion. The nucleus is more stable even with the additional Coulomb
"unbinding energy" because the nuclear effects "trump" the Coulomb effects.
So the extra energy the neutron brings in doesn't have to be enough to get out of the
"old" potential well of the target nucleus - it has to be enough to get out of the well of
the "new" nucleus.
You really have to due a quantum mechanical description of the nucleus with all the
shell structure; just like one has to do with electrons in an atom.
Yes - also discovered that if you make a mistake in a "tex" equation, and you edit it;
the system doesn't re-evaluate the tex expression after you edit it. You have to copy
the entire post to an editor [ I use Emacs instead of Notepad because I'm working in
Unix], delete the faulty post, and post a new reply using the saved text from the editor.
Dr. Gregory Greenman