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ionisation energy of He is 19.6 xx 1...

ionisation energy of He is `19.6 xx 10^(-18) j atom^(-1 ) ` The energy of the first stationary state (n=1) of `Li^(+)` is

A

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

B

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

C

`-2.20xx10^(-15)J atom^(-1)`

D

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

Text Solution

Verified by Experts

The correct Answer is:
B
`E_(1)/E_(2)=Z_(1)^(2)/Z_(2)^(2)xxn_(2)^(2)/n_(1)^(2)`
`((-19.6xx10^(-18)))/(E_(2))=2^(2)/3^(2)xx1^(2)/1^(2)`
`E_(2)=-4.41xx10^(-17)J atom^(-1)`
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