# An electron in a mercury atom jumps from level a to level g by absorbing a single photon. How do you determine the energy of the photon in joules?

May 22, 2017

Relativistic effects are present here... but I assume that the transition from level $a$ to level $g$:

• has $\Delta L = \pm 1$ (the total change in angular momentum is exactly $1$).
• has $\Delta S = 0$ (no changing between spin-paired and spin-parallel).
• is not influenced by relativistic effects (e.g. those that make one transition more favorable than another in cases of close-lying atomic orbitals, particularly big $f$ orbitals).

If that is the case, then since we only care about the photon...

$\Delta E = h \nu$

...as long as you know the frequency of whatever light you chose in ${\text{s}}^{- 1}$. $h$ is Planck's constant, $6.626 \times {10}^{- 34} \text{J"cdot"s}$.

I will not use the Rydberg equation, as it only works on hydrogen atom. Clearly, mercury has more than one electron, so it has nondegenerate orbitals of the same $n$ with differing $l$ (unlike hydrogen atom).