Question

Find the Rydberg correction for the `3P` term of a `Na` atom whose first exitation potential is `2.10V` and whose valance electron in the normal `3S` state has the binding energy `5.14 eV`.

Answer 1

The enrgy of the `3p` state must be `-(E_(0)-e varphi)` where `-E_(0)` is the energy of the `3S` state. Then
`E_(0)-evarphi_(1)=( ħR)/((3+alpha_(1))^(2))`
so `alpha_(1)=sqrt(( ħR)/(E_(0)-e varphi_(1)))-3 =-0.885`