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

1 Answer
May 22, 2017

Relativistic effects are present here... but I assume that the transition from level aa to level gg:

  • has DeltaL = pm 1 (the total change in angular momentum is exactly 1).
  • has DeltaS = 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...

DeltaE = hnu

...as long as you know the frequency of whatever light you chose in "s"^(-1). h is Planck's constant, 6.626 xx 10^(-34) "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).