Chemistry - Atomic Structure Question with Solution | TestHub

For H atom & H like species
The electronic energy in the nth orbit is

${\text{E}}_{\text{n}}=-\text{13.6}\frac{{\text{z}}^2}{{\text{n}}^2}eV$

${\left({\text{E}}_{\text{n}}\right)}_{{\text{Li}}^{2+}}=\frac{-\text{13.6}\times 9}{{\text{n}}^2}eV=\frac{-\text{122.4 eV}}{{\text{n}}^2}$

If n = 2

${\left({\text{E}}_2\right)}_{{\text{Li}}^{2+}}=\frac{-\text{122.4}}{{\text{2}}^2}=-\text{30.6 eV}$