1
$\begingroup$

For a given isotope, does the nuclear binding energy per nucleon depend on the presence of electrons? For instance, if an electron was excited by an incoming photon and jumps to a different orbital, will the nuclear binding energy change?

$\endgroup$
  • $\begingroup$ Yes, but very weakly. There are tiny chemical modulations of nuclear energies, so in theory nuclear decays and reactions are susceptible to chemical composition, temperature, pressure etc.. but the effects are so small that they can be neglected for all purposes as far as I know. $\endgroup$ – CuriousOne May 25 '15 at 1:41
  • $\begingroup$ Like what CuriousOne said, yes. Think of hydrogen. You've got one nucleon, a proton. You've also got an electron. The binding energy is -13.6keV. An incoming photon comes along, the electron jumps to a different orbital, and the binding energy has changed. $\endgroup$ – John Duffield May 25 '15 at 7:39
  • $\begingroup$ Why would hydrogen have a binding energy? en.wikipedia.org/wiki/Hydrogen_atom Wikipedia says the binding energy of hydrogen is exactly 0? $\endgroup$ – Joshua Lin May 25 '15 at 7:41
  • $\begingroup$ @JoshuaLin The nuclear binding energy of hydrogen is zero. The hydrogen atom in its ground state has a binding energy of 13.6 eV. It is bound because it has a binding energy. Can you be clearer in your question about what definition of binding energy you want to know about. $\endgroup$ – Rob Jeffries May 25 '15 at 8:46
  • $\begingroup$ Sorry, I meant the nuclear binding energy $\endgroup$ – Joshua Lin May 25 '15 at 8:48
1
$\begingroup$

In an answer to another question I made reasonable estimate of the amount of electron charge that's contained within the nucleus. (Actually for that question I used a volume somewhat larger than the nucleus, but the method is there.) To first approximation, all the electronic charge within the nuclear volume comes from the two $1s$ electrons. The distribution of a full $p$-wave or higher shell is spherically symmetric. By the shell theorem, a spherically symmetric shell of charge does not affect the energy levels of systems inside.

Since the nuclear interaction is mostly blind to electric charge, and the charge within the nucleus is always a tiny correction to the proton charge, the effect on nuclear structure due to electronic charges must be very small.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.