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?

Md Ariful Islam

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Md Ariful Islam

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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 , $6.626 xx 10^(-34) "J"cdot"s"$.

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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