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

Explanation

Energy of an hydrogen like atom like atoms `He^(+)` in an `n^("th")` orbit is given by `E_(n)=-(13.6Z^(2))/(n^(2))eV`
For hydrogen atom, `Z=1 therefore E_(n)=-(13.6)/(n^(2))eV`
For ground state, `n=1 therefore E_(1)=-(13.6)/(1^(2))eV=-13.6eV`
For `He^(+)` ion, `Z=2`
`E_(n)=-(4(13.6))/(n^(2))eV`
For first excited state, `n=2 therefore E_(2)=-(4(13.6))/((2)^(2))eV=-13.6eV`