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Ionization energy of He^(+) is 19.6 xx10...

Ionization energy of `He^(+)` is `19.6 xx10^(-18)" J atom"^(-1)` The energy of the first stationary state (n=1) of `Li^(2+)` is :

A

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

B

`8.82 xx 10^(-17)"J atom"^(-1)`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
D
`(I_(He^(+)))/(I_(Li^(2+)))=(Z_(1)^(2))/(Z_(2)^(2)), (19.6xx10^(-18))/(I_(Li^(2+))) =(4)/(9)` `I_(Li^(2+))=(9)/(4) xx 19.6 xx 10^(-18)=44.1 xx 10^(-18)=4.41 xx 10^(-17)"J atom"^(-1)" " E_(Li^(2+))= -4.41 xx 10^(-17)"J atom"^(-1)`
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