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

`E_(n) = - (2pi^(2) m Z^(2) e^(4))/(n^(2) h^(2)) = - K (Z^(2))/(n^(2))`
I.E. of `H^(+) = E_(oo) - E_(1) = 0 - (-K (2^(2))/(1^(2)))`
`= 4K = 19.6 xx 10^(-18) J "atom"^(-1)` (Given)
or `K = 4.9 xx 10^(-18) J "atom"^(-1)`
For `Li^(2+) , Z = 3` and for first stationary state, n = 1
`:. E_(1) = - K (Z^(2))/(n^(2)) = - 4.9 xx 10^(-18) xx (3^(2))/(1^(2))`
`= -4.41 xx 10^(-17) J "atom"^(-1)`