The highest excited state that unexcited hydrogen atom can ereach when they are bombarged with ` 12. 2 eV` electron is :

Options

A. ` n=1`
B. ` n=2`
C. ` n=3`
D. ` n=4`

Answer

C

Explanation

`E_(1) = - 13.6eV`
After absorpton fo ` 12. 2 eV` energy
` E_(H) = - 13 . 6 + 12 . 2 =- 14 eV`
Now ` E_(n) = E_(t) /n^(2) :. N^(2) = ( - 13 . 6 )/( -1.4) = 9.71`
`therefore n=3` .