The binding energy of the valance electron in a `Li` atom in the states `2S` and `2P` is equal to `5.39` and `3.54 eV` respectively. Find the Rydberg corrections for `s` and `p` terms of the atom.

The binding energy of the valance electron in a `Li` atom in the state

1.	The binding energy of the valance electron in a `Li` atom in the states `2S` and `2P` is equal to `5.39` and `3.54 eV` respectively. Find the Rydberg corrections for `s` and `p` terms of the atom.
Answer» From the Rydberg formula we write `E_(n)= -( ħR)/((n+alpha_(l)^(2))` we use ` ħR= 13.6eV`. Then for n= 2 state `5.39= -(13.6)/((2+alpha_(0))^(2)), l=0(S)` state `alpha_(0)~~ -0.41` for `p` state `3.54= -(13.6)/((2+alpha_(1))^(2))` `alpha_(1)=-0.039`.

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The binding energy of the valance electron in a `Li` atom in the states `2S` and `2P` is equal to `5.39` and `3.54 eV` respectively. Find the Rydberg corrections for `s` and `p` terms of the atom.

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