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

D

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

Text Solution

Verified by Experts

The correct Answer is:
C
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)`
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