15 eV is given to `e^(-)` in `4^("th")` orbit then find it's final energy when it comes out of H-atom

Options

A. 14.14 eV
B. 13.6 eV
C. 12.08 eV
D. 15.85

Answer

A

Explanation

Energy of `4^("th")` orbit of H-atom `=-13.6xx(1)/(16)=-0.85eV`
S energy release = total - ionization energy of `4^("th")` orbit
`=15-0.85`
`=14.15`