The photon emitted due to electronic transition of `5^(th)` excited state to `2^(nd)` excited state in `Li^(2+)`, is used to excite `He^+` already in first excited state.`He^+` ion after absorbing the photon reaches in an orbit having total energy equal to :

Options

A. `-3.4 eV`
B. `-13.6 eV`
C. `-6.8 eV`
D. `-27.2 eV`

Answer

A

Explanation

(A)For `Li^(2+)`, n=6 to n=3
For H, the similar transition is 2 to 1
For `He^+` , the similar transition is 4 to 2
Energy of `4^(th)` orbit of `He^(+)` =`-13.6xx2^2/4^2= -3.4 eV`