Question

Determine the wavelength of spectral lines appearing on transition of exicted `Li` atoms from the state `3S` down to the ground state `2S`. The Rudberg corrections for the `S` and `P` terms are `-0.41` and `-0.04`.

Answer 1

The enrgy of the `3S` state is
`E(3S)=-(ħR)/((3-0.41)^(2))=-2.03eV`
The energy of a `2S` state is
`E(2S)= -(ħR)/((2-0.41)^(2))= -5.39eV`
The energy of a `2P` state is
`E(2P)=(ħR)/((2-.04)^(2))=-3.55eV`
We see that
`E(2S) lt E(2P) lt E(3S)`
The transitions are `3S rarr2P` and `2Prarr2S`.
Direct `3Srarr2S` transition is forbidden by selection rules. The wavelength are determined by
`E_(2)-E_(1)= DeltaE=(2piħc)/(lambda)`
Substitution gives
`lambda= 0.816mu m(2Srarr2P)`
and `lambda=0.67mu m(2Prarr2S)`