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Ionisation energy of He^+ is 19.6 xx 10^...

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.

A

`8.82xx10^(-17) "J atom"^(-1)`

B

`4.41xx10^(-16) "J atom"^(-1)`

C

`-4.41xx10^(-16) "J atom"^(-1)`

D

`-2.2xx10^(-15) "J atom"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
`E_n=-13.6(Z^2)/n^2 eV`
`E prop Z^2`
IP of `He^(+) =-E_1 = 19.6xx10^(-18) "J atom"^(-1)` (given)
`E_1` for `Li^(+2) = ?`
`(E_(He^(+)))/(E_(Li^(2+)))=((2)^2)/((3)^2)=4/9`
`E_(Li^(+2))=9/4(-19.6 xx10^(-18))=-44.1 xx10^(-18)`
`=-4.41 xx10^(-17) " J atom"^(-1)`
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