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 of a h- like atom in it.s `n^(th)` state is given by
` E_n =- Z^2 xx ( 13.6 )/( n^2 ) eV `
for first excited state of ` he ^(+ ) , n =2 ,z =2`
` therefore E _( Hr^(+)) =- ( 4 )/(2^2) xx13.6 =- 13.6 eV `