For each choice of $n$ and $l$ (with $l \neq 0$ ), a nucleon has two different possible energy levels with $j=l \pm \frac{1}{2}$. Prove that the sum of the degeneracies of these two levels is equal to the total degeneracy that the level $n l$ would have had in the absence of any $\mathbf{S} \cdot \mathbf{L}$ splitting.

The ground state of ${ }^7 \mathrm{Li}$ has $j_{\text {tot }}=\frac{3}{2}$ and its first excited state has $j=\frac{1}{2}$. Explain these observations in terms of the shell model.

Use the shell model to predict $j_{\text {tot }}$ for the ground states of ${ }_{19}^{39} \mathrm{~K}_{20},{ }_{20}^{40} \mathrm{Ca}_{20}$, and ${ }_{21}^{41} \mathrm{Sc}_{20}$.

Apply Appendix D to calculate the chemical symbols and the natural percent abundances of the isotopes with (a) $Z=2, \quad A=3 ; \quad$ (b) $Z=6, \quad A=13$; (c) $Z=26, A=56$; (d) $Z=82, A=206$.Appendix D to find the chemical symbols and the half-lives of the four radioactive nuclei with (a) $Z=1, \quad A=3 ;$ (b) $Z=6, \quad A=14$; (c) $Z=19, A=40$; (d) $Z=94, A=244$.Appendix $\mathrm{D}$ to make a list of all the stable isotopes, isobars, and isotones of ${ }^{70} \mathrm{Ge}$. (b) Do the same for ${ }^{27} \mathrm{Al}$ and (c) ${ }^{88} \mathrm{Sr}$.

Use the levels shown in the above figure to predict the total angular momenta of the following nuclei: ${ }^{29} \mathrm{Si}$, ${ }^{33} \mathrm{~S},{ }^{37} \mathrm{Cl},{ }^{59} \mathrm{Co}$, and ${ }^{71} \mathrm{Ga}$.

Use the levels shown in the above figure to predict $j_{\text {tot }}$ for the ground states of the nine nuclei ${ }^{40} \mathrm{Ca},{ }^{41} \mathrm{Ca}$, ${ }^{48} \mathrm{Ca}$. Compare with the data in Appendix D.

Which of the following nuclei has a closed subshell for Z? Which for $N ?{ }^{10} \mathrm{~B},{ }^{11} \mathrm{~B},{ }^{17} \mathrm{O},{ }^{27} \mathrm{Al},{ }^{28} \mathrm{Si},{ }^{42} \mathrm{Ca},{ }^{48} \mathrm{Ca}$.

(a) Substitute the binding-energy formula into the relation

$$m_{\mathrm{nuc}}=Z m_{\mathrm{p}}+N m_{\mathrm{n}}-\frac{B}{c^2}$$

to obtain the semiempirical mass formula. [To simplify matters, take $A$ to be odd, so that the pairing term in above is zero.] (b) Among any set of isobars, the nucleus with lowest mass is the most stable. To identify this nucleus, first write your mass formula in terms of the variables $A$ and $Z$ (that is, replace $N$, wherever it appears, by $A-Z$ ). With $A$ fixed, differentiate $m$ with respect to $Z$. The minimum mass is determined by the condition $\partial m / \partial Z=0$. Show that this leads to a relation of the form

$$Z=\frac{A}{2} \frac{1+\alpha}{1+\beta A^{2 / 3}}$$

where $\alpha$ and $\beta$ are related to the coefficients as follows:

$$\alpha=\frac{\left(m_{\mathrm{n}}-m_{\mathrm{p}}\right) c^2}{4 a_{\mathrm{sym}}} \quad \text { and } \quad \beta=\frac{a_{\mathrm{Coul}}}{4 a_{\mathrm{sym}}}$$

(c) For $A=37,115,185$, find the values of $Z$ that give the most stable nuclei and compare with the observed values from Appendix D.

In this problem you will calculate the electrostatic energy of a uniform sphere of charge $Q$ with radius $R$. The electric potential $V(r)$ at any radius $r<R$ is given by the equation in the above problem, and the potential energy of a charge $q$ at radius $r$ is $q V(r)$. (a) Write down the total charge contained between radius $r$ and $r+d r$, and hence find the potential energy of that charge. (b) If you integrate your answer to (a) from $r=0$ to $r=R$, you will get twice the total potential energy of the whole sphere since you will have counted the energy of any two charge elements twice. Show that the total Coulomb energy of the sphere is

$$U_{\text {Coul }}=\frac{3}{5} \frac{k Q^2}{R}$$

Compare the surface area of a sphere to that of a cube of the same volume. How much energy, in $\mathrm{MeV}$, would be needed to distort the normally spherical ${ }^{60} \mathrm{Ni}$ nucleus into a cubical shape if the volume and Coulomb energies changed by negligible amounts?

If ${ }^{235} \mathrm{U}$ captures a neutron to form ${ }^{236} \mathrm{U}$ in its ground state, the energy released is $B\left({ }^{236} \mathrm{U}\right)-B\left({ }^{235} \mathrm{U}\right)$. (a) Prove this statement. (b) Use the binding-energy formula to estimate the energy released, and compare with the observed value of $6.5 \mathrm{MeV}$. (Note: We have assumed here that ${ }^{236} \mathrm{U}$ is formed in its ground state, and the $6.5 \mathrm{MeV}$ is carried away, by a photon, for example. An important alternative is that ${ }^{236} \mathrm{U}$ can be formed in an excited state, $6.5 \mathrm{MeV}$ above the ground state. This excitation energy can lead to oscillations that cause the nucleus to fission.)

Use the binding-energy formula to calculate the total binding energies of ${ }_{26}^5 \mathrm{Ni},{ }_{27}^{56} \mathrm{Co},{ }_{26}^{56} \mathrm{Fe}$ and ${ }_{25}^{56} \mathrm{Mn}$. Which terms are different in the four cases, and why?

Use the semiempirical binding-energy formula to find $B / A$ for ${ }^{20} \mathrm{Ne},{ }^{60} \mathrm{Ni},{ }^{259} \mathrm{No}$. Make a table showing all five terms for each nucleus, and comment on their relative importance. Do your final answers follow the general trend of the above figure?

(a) Use the binding-energy formula to predict the binding energy of ${ }^{40} \mathrm{Ca}$. (b) Compare your answer with the actual binding energy found from the masses listed in Appendix D.