The energy of a hydrogen atom in the ground state is `-13.6 eV`. The eneergy of a `He^(+)` ion in the first excited state will be

Options

A. `-13.6 eV`
B. `-27.2 eV`
C. `-54.4 eV`
D. `-6.8 eV`

Answer

A

Explanation

Energy `E` of an atom with principle quantum number `n` is given by `E = (-13.6)/(n^(2)) Z^(2)` for first excited state `n = 2` and for `He^(+) Z = 2`
`rArr E = (-13.6 xx (2)^(2))/((2)^(2)) =- 13.6 eV`