The energy of an electron in the first Bohr orbit of H atom is `-13.6 eV` The potential energy value (s) of excited state(s) for the electron in the Bohr orbit of hydrogen is(are)

Options

A. `-3.4 eV`
B. `-4.2 eV`
C. `-6.8 eV`
D. `+6.8 eV`

Answer

A

Explanation

`E_(2)=-(13.6 eV)/(n^(2))`
For `n=2, E_(2)=-(13.6eV)/(4)=-3.4 eV`.