The larger the total binding energy of a nucleus, the greater the minimum energy that must be added to infinitely separate the constituent nucleons. Shouldn't the total binding energy, then, be used to determine the stability of a nucleus? I don't understand why the binding energy per nucleon is used instead.
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Because in a system of $n$ nucleons you look for combinations of nucleons that have the highest binding energy taken all together. It is not a one-body problem, it is an $n$-body problem. $\endgroup$– Jon CusterCommented May 4, 2022 at 17:47
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Jon Custer is right, here is why.
The total binding energy is the sum of the energy in the attractive part due to the residual nuclear force between quarks in neighboring nucleons and the repulsive part due to the proton-proton electrostatic repulsion.
The attractive part is strictly local, acting to stick protons and neutrons together that are close enough to touch- it is a very short-ranged force. On the other hand, a single proton in the nucleus feels the repulsion of all the other protons in the nucleus, not just its nearest neighbors. Once the net attractive contribution becomes equal to the repulsive contribution, the binding energy per nucleon approaches zero and there's almost nothing left to make it energetically favorable for the nucleus to remain in one piece: give it just the right bump, and it flies apart.
This is why the binding energy per nucleon is so strong in helium and relatively much weaker in uranium.
answered May 4, 2022 at 19:04