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

Options

A. `=13.6eV`
B. `gt13.6eV`
C. `lt13.6eVandgt3.4eV`
D. `le3.4eV`

Answer

D

Explanation

`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`.