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The energy required to dislodge electron...

The energy required to dislodge electron from excited isolated H-atom `(IE_(1) = 13.6 eV)` is

A

`=13.6eV`

B

`gt13.6eV`

C

`lt13.6eVandgt3.4eV`

D

`le3.4eV`

Text Solution

Verified by Experts

The correct Answer is:
D
`IE_(1)=E_(1)-E_(oo)`
`13.6=a(1/n_(1)^(2)-1/n_(2)^(2))" "(.a."is a constant")`
`13.6=a(1/1^(2)-1/oo)`
a = 13.6
If the excited state is `n_(1)` = 2 then
`DeltaE=a(1/n_(1)^(2)-1/n_(2)^(2))eV=3.4eV`
Value of `DeltaE` will be less than 3.4 eV for greater values of `n_(1)` i.e. when `n_(1)gt2`
So `DeltaEle3.4eV`.
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