The energy of the electron in the ground state of H-atom is `-13.6 eV`. The energy of the first excited state will be

Options

A. `-3.4 eV`
B. `-27.2 eV`
C. `-6.8 eV`
D. `-52.4 eV`

Answer

A

Explanation

`E_(n)=E_(1)/n^(2)`
Thus, energy of electron in the first excited state is given by,
`E_(2)=E_(1)/4=(-13.6)/4=-3.4 eV`