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?

1 Answer
Truong-Son N.
May 22, 2017

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

  • 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).

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