The ionisation energy of `He^+" is "19.6 xx 10^(-12)` J/atom. Calculate the energy of the first stationary state of `Li^(2+)`.

Options

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

Answer

C

Explanation

Ionisation energy of `He^(+)=19.6xx10^(-18)"J atom"^(-1)`
= Energy of first stationary state of `He^(+)`
`=-19.6xx10^(-18)" J atom"^(-1)." "E_(n(He^(+)))=(-2pi^(2)Z_(He)^(2)me^(4))/(n^(2)h^(2)),E_(n(Li^(+))2pi^(2)Z_(Li)^(2)me^(4))/(n^(2)h^(2))`
`(E_(n(Li^(2+))))/(E_(n(He^(+))))=(Z_(Li)^(2))/(Z_(He)^(2+))=(3^(2))/(2^(2))" or, "E_(1)(Li^(2+))=(3^(2))/(2^(2))E_(1(He^(+)))=(9)/(4)(-19.6xx10^(-18))`
`=-4.41xx10^(-17)"J atom"^(-1)`