Ionization energy of `He^(+)` is `19.6 xx10^(-18)" J atom"^(-1)` The energy of the first stationary state (n=1) of `Li^(2+)` is :

Options

A. `-2.2xx10^(-15)" J atom"^(-1)`
B. `8.82xx10^(-17)" J atom"^(-1)`
C. `4.41xx10^(-16)" J atom"^(-1)`
D. `-4.41xx10^(-17)" J atom"^(-1)`

Answer

D

Explanation

`(I_(He^(+)))/(I_(Li^(3+)))=(Z_(1)^(2))/(Z_(2)^(2)), (19.6xx10^(-18))/(I_(L^(3+)))=(4)/(9)`
`I_(Li^(3+))=(9)/(4)xx19.6xx10^(-18) =44.1xx10^(-18)=4.41xx10^(-17)" J atom"^(-1)`
`E_(Li^(3+))=-4.41xx10^(-17" J atom"^(-1)`