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The IE(1) of Li is 5.4 eV and IE(1) of H...

The `IE_(1)` of `Li` is `5.4 eV` and `IE_(1)` of `H` is `13.6 eV`. Calculate the charge acting on the outermost electron of `Li` atom.

Text Solution

Verified by Experts

Electronic configuration of `Li` is `1s^(2) 2s^(1) (n=2)`.
`E_(Li) = ((Z^(2))_(Li))/(n^(2))xx E_(H) rArr Z_(Li) = nsqrt((E_(Li))/(E_(H)))`
`= 2xx((5.4)/(13.6))^((1)/(2)) = 1.26 eV`
`Z_("eff")` of `Li` is `1.26 eV` because `2s` electrons are shielded by `1s^(2)` electrons.
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