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.

Answer

`-4.41xx10^(-17) J "atom"^(-1)`

Explanation

lonisation energy of `He^(+)` ion =`19.6xx10^(-18) J "atom"^(-1)`
Energy of first orbit (ground state) of `He^(+)` ion =E_(prop)-l.E.`=0-19.6xx10^(-18)=-19.6xx10^(-18) J "atom"^(-1)`
Energy of first stationary state (ground state) o f`Li^(+)` ion =`(-19.6xx10^(-18)xx(Z_(Li))^(2))/((Z_(He))^(2))`
`=-19.6xx10^(-18)xx(9)/(4)=-44.1xx10^(-18)J "atom"^(-1)=-4.41xx10^(-17) J "atom"^(-1)`.