A hydrogen atom in state `n = 6` makes two successive transition and reaches the ground state in the first transition a photon of `1.13eV` is emitted (a) Find the energy of the photon emitted in the second transition (b) what is the value of n in the intermediate state?

Answer

A, B, C

Explanation

`n=6`, `E= (-13.6)/(36) = -0.3777777`
Energy in grounds tate`=-13.6 eV`.
Energy emitted in Second transition
`=-13.6 -(0.3777bar7 + 1.13)`
`=-12.09 = 12.1 eV`.
(b) Energy in the intermediate state
`=1.13 eV + 0.37`
implies `(18.6 xx Z^2)/(n^2) = 13.0/n^2 = 1507bar7`
or `n=sqrt((13.6)/(1.507)) = 3.03 =3`