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?
Relativistic effects are present here... but I assume that the transition from level
#DeltaL = pm 1#(the total change in angular momentum is exactly #1#).
#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
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
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