The `IE_(1)` of `H` is `13.6 eV`. It is exposed to electromagnetic waves of `1028Å` and gives out induced radiation. Find out orbit of these induced radiation.

Answer

3

Explanation

`E_(1)` for H-atom `=-13.6`
`:' E=12375/lambda`, when `lambda` is in `Å`
`:.` Energy given to H-atom `=12375/1028 eV=12.07 eV`
`:.` Energy of H-atom after excitation `=-13.6+12.07=-1.53 eV`
`:' E_(n)=E_(1)/n^(2)`
`:. N^(2)=(-13.6)/(-1.53)=9`,
`:. n=3`