The energy of an electron in first Bohr's orbit of H atom is `-13.6eV`. The energy value of electron in the first excited state of `Li^(2+)` is :

Options

A. `-27.2 eV`
B. `30.6ev`
C. `-30.6 eV`
D. `27.2 eV`

Answer

D

Explanation

For `Li^(2+)` ion
`E = - 13.6 xx (Z^(2))/(n^(2)) = - 13.6 xx ((3)^(2))/((2)^(2))`
` = (- 13.6 xx 9)/(4) = - 30.6 eV`