Sodium Energy and Angular Momentum of Electrons In a sodium atom, why do the $s$ and $p$  angular momentum states drop to the $3p$  rather than straight to $3s$ ? (See the energy level diagram below.)

 A: A photon carries angular momentum $J=1$ (in units of $\hbar$). This results in what are known as "selection rules" for transitions between atomic states.  The most important one is that the strongest transitions are those in which the atomic angular momentum $L$ changes by at most one unit, $L\rightarrow L-1$, $L$, or $L+1$, because the photon has to carry away one unit of angular momentum.  These are known as "allowed" or "electric dipole" transitions.  (The "electric dipole" comes from the fact that the electric dipole moment of the atomic wave function changes when the photon is spontaneously emitted.)
The photon angular momentum can be oriented in any direction; it is ultimately determined by the polarization and propagation direction of the photon.  It should be easy to see how this can lead to $L\rightarrow L-1$ or $L\rightarrow L+1$ transitions, with the photon carrying one unit of angular momentum either parallel or antiparallel to the atomic $\vec{L}$.  However, when the atomic $\vec{L}$ and photon $\vec{J}$ are oriented along different directions, it is possible to leave the magnitude $L$ unchanged.  (For $L=1$, think of the atomic angular momentum $\vec{L}$, the outgoing photon angular momentum $\vec{J}$, and the final atomic angular momentum $\vec{L}-\vec{J}$ as forming an equilateral triangle.)  It isn't possible to have $L\rightarrow L$ transitions when $L=0$, however; there's no way to draw a triangle like I described if $\vec{L}$ is just the zero vector.
There are higher-order spontaneous transitions possible as well (electric quadrupole, magnetic dipole, etc.).  These are possible because the outgoing photon can itself carry orbital angular momentum, but they are not usually seen, unless an electric dipole transition is, for some reason, not possible.  The inclusion of photon orbital angular momentum ensures that the quantum mechanical matrix elements (and thus rates) for these are generally suppressed by powers of the fine structure constant $e^{2}/4\pi\epsilon_{0}\hbar c=1/137$.  However, even with the inclusion of higher multipoles, $0\rightarrow0$ transitions are still not permitted.  (Actually $0\rightarrow0$ spontaneous decays, like the decay of the $2S$ state of hydrogen, involve the simultaneous emission of two photons to get around the angular momentum restrictions. This makes the decays very slow; the hydrogen $2S$ state lives, on average, about $1/7$ of a second.)
Finally, transitions generally go faster when the photon energy is larger.  The rate for dipole transitions goes like $\omega^{3}$.  So what that figure shows is the predominant spontaneous decay mode for each atomic state.  The state decays to the lowest-energy state that is permitted by the electric dipole selection rule:  $L\rightarrow L-1$, $L$, or $L+1$; no $0\rightarrow0$.  The $P$ states all decay directly to the ground state ($3S$), but the $S$ states cannot decay to another $S$ state, so they go to the next-lowest state, $3P$.  The higher angular momentum states ($D$ and $F$, with $L\geq2$) cannot go to $3S$ either, so they go to the lowest state with $L\rightarrow L-1$.
