For the Franck-Hertz experiment there is a voltage drop at $4.9\rm\,V$. What transition does this represent in the mercury? Looking at the energy levels it seems to be from the ground to the 2nd excited state. But would this state not be occupied by other electrons? Any references would also be great.
Answer
According to a list of levels from NIST, the ground state for mercury has quantum numbers $^1S_0$ (with the electron in the 6th shell). I usually have to look up how to read those symbols: the $^1$ tells you it's a spin singlet, the $S$ tells you that the orbital angular momentum $L=0$, and the $_0$ tells you the value of the total angular momentum quantum number $J$. The first excited states are a $^3P$ triplet (spin triplet, $L=1$) with \begin{align} \begin{array}{cc} J & \text{Wavenumber }\mathrm{(cm^{-1})} & \text{Energy (eV)} \\ \hline 0 & 37\,600 & 4.66 \\ 1 & 39\,400 & 4.88 \\ 2 & 44\,000 & 5.45 \end{array} \end{align} The selection rules for atomic transitions forbid transitions where $J$ goes from zero to zero. In photon-mediated transitions this is pretty easy to understand: the photon must carry away one unit of angular momentum. (You can have transitions with $\Delta J=0$ for nonzero $J$; in that case you can imagine that the orientation $m_J$ of the atom's spin must change.) Apparently the same rules apply to the collisional transitions seen in the Franck-Hertz experiment — otherwise you could excited the $^3P_0$ first excited state and the energy involved would be $4.7\rm\,V$. Very interesting observation!
In your question you ask,
would this state not be occupied by other electrons?
Here the answer is no. In that same list of levels you can see that the ground state configuration is $5d^{10}({}^1S)6s^2$, while the first excited states have $5d^{10}({}^1S)6s6p$ (on top of a xenon-like core of 54 inert electrons).
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