Consider two levels of the helium atom in both of which the spins are antiparallel and one electron is in an $s$ state $(l=0)$. In the higher level the second electron occupies a $d$ state $(l=2)$, and in the lower level it occupies a $p$ state $(l=1)$. (a) Sketch the splitting of both levels resulting from a magnetic field along the $z$ axis. (b) Imagine a transition from one of the $d$ states, with $L_{z}=m_{i} \hbar$, to one of the $p$ states, with $L_{z}=m_{f} \hbar$. Since $m_{i}$ can be $2,1,0,-1$, or $-2$ and $m_{f}$ can be 1,0 , or $-1$, there are $5 \times 3$ or 15 distinct conceivable transitions. How many different photon energies would these 15 transitions produce? (c) Not all of these 15 transitions occur. In fact, it is found that the only transitions observed are those for which
$$
\left(m_{f}-m_{i}\right)=1, \quad \text { or } 0, \quad \text { or }-1 \quad(9.36)
$$
(A restriction like this on the transitions that take place is called a selection rule, as we discuss in Chapter 11.) Prove that because of the restriction (9.36), there are only three distinct photon energies produced in all possible transitions. (This means that the normal Zeeman effect always produces just three spectral lines, however large the angular momenta involved.)