The photoelectron spectrum of krypton is shown below.enter image description here

(where the x-axis is in electron volts (eV))\

Since both the $3p$ and $3d$ shells are more than half full, based on Hund's rules the lowest-energy level is the one with the largest J, i.e. the $2\mathrm{P}_{3/2}$ and $2\mathrm{D}_{5/2}$ states should be lowest in energy (higher value in eV). However, based on the peak intensities, this is not the ordering observed (see labeling in diagram). Why is this the case here?

  • $\begingroup$ I think lowest J is the lowest energy, see here for example. $\endgroup$ – ron May 21 '15 at 14:08
  • 1
    $\begingroup$ This is only true for shells that are less than half full. In Krypton's case, the 3p and 3d shells are full so largest J values are lowest in energy. $\endgroup$ – 218 May 21 '15 at 14:27

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