Answer
See answers.
Work Step by Step
See Example 30-4.
$^{23}_{11}Na$ has 11 protons and 12 neutrons. Calculate the binding energy using the masses of the components and the mass of the nucleus. See Appendix B.
$$E_{binding}=\left( 11m(^{1}_{1}H)+12m(^{1}_{0}n)-m(^{23}_{11}Na)\right)c^2$$
$$ =\left( 11(1.007825u)+12(1.008665u)-(22.989769u)\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$ =186.6MeV$$
Binding energy per nucleon is 186.564MeV/23 nucleons = 8.111MeV/nucleon.
$^{24}_{11}Na$ has 11 protons and 13 neutrons. Calculate the binding energy using the masses of the components and the mass of the nucleus. See Appendix B.
$$E_{binding}=\left( 11m(^{1}_{1}H)+13m(^{1}_{0}n)-m(^{24}_{11}Na)\right)c^2$$
$$ =\left( 11(1.007825u)+13(1.008665u)-(23.990963u)\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$ =193.5MeV$$
Binding energy per nucleon is 193.534MeV/24 nucleons = 8.063MeV/nucleon.
We see that the nucleons in Na-23 are more tightly bound than those in Na-24. We expect Na-23 to be more stable than Na-24, which is the case.
