Ionisation energy of `He^+` is `19.6 xx 10^-18 J "atom"^(-1)`. The energy of the first stationary state `(n = 1)` of `Li^( 2 +)` is.

Options

A. `-2.2 xx 10^(-15) J "atom"^(-1)`
B. `8.82 xx 10^(-17) J "atom"^(-1)`
C. `4.41 xx 10^(-16) J "atom"^(-1)`
D. `-4.41 xx 10^(-17) J "atom"^(-1)`

Answer

D

Explanation

IE of `He^(+)` ion = IE of H atom `xx Z^2`
= I.E. of H atom `xx (2)^(2)`
`19.6 xx 10^(-18) J "atom"^(-1) = 4 xx` I.E. of H atom
`:.` I.E. of H atom = `(19.6 xx 10^(-18))/(4)`
`= 4.9 xx 10^(-18) J "atom"^(-1)`
I.E. of `Li^(2+)` ion = I.E. of H atom `xx (3)^2`
I.E. of `Li^(2+)` ion = `4.9 xx 10^(-18) xx 9`
`= 4.41 xx 10^(-17) J`
Energy = `-4.41 xx 10^(-17) J`