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

Answer

A

Explanation

`E_(He^(+))=-(Z^(2))/(n^(2))xx13.6eV`
for `He^(+)` ion, Z=2, and for first excited state, n=3.
`:. E_(He^(+))=-(2^(2))/(2^(2))xx13.6eV=-13.6 eV`