The ionisation energy of `He^(o+)` is `19.6 xx 10^(-18) J "atom" ^(-1)` .The energy of the first stationary state of `Li^(2+)` will be

Options

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

Answer

2

Explanation

I.E. of `He^(+)=19.6xx10^(-18)J atom^(-1)`
I.E. `=-E_(1)`
`E_(1)` for `He^(+)` is `=-19.6xx10^(-18) J atom^(-1)`
`((E_(1))_(He^(+)))/((E_(1))_(Li^(3+)))=((Z_(He^(+)))^(2))/((Z_(Li^(2-)))^(2))rArr (-19.6xx10^(-18))/((E_(1))_(Li^(2-)))=4/9`
`E_(1)(Li^(2+))=(-19.6xx9xx10^(-18))/(4)=-44.1xx10^(-18)=-4.41xx`0^(-17) J atom^(-1)`