Electronic of energy `12.09` eV can excite hydrogen atoms . To which orbit is the electron in the hydrogen atom raised and what are the wavelenths of the radiations emitted as it drops back to the ground state ?

Explanation

The energies of the electron in different are:
`E _(1) = - 13.6 ` for `n =1`
`E (2) = 3.4 eV` for `n =2`
`and E _(3) = - 1. 51 eV`for `n =3`
Evidently, the energy ennded by an electron to go to the `E _(3)` level `(n = 3 or ` M-level) is `13.6 - 1. 51 = 12. 09 eV.` Thus the electro is raised to the third orbit of principle quantum number `n =3.`
Now, an electron in the `n=3` level can return to the ground state by making the following possible jups,
(i) `n = 3` to `n =2` and then from `n =2` to `n =1.`
(i) `n = 3` to `n =1.`
Thus the corresponding wavelengths emitted are:
(a) For n =3 to `n =2,`
`(1)/(lamda _(1)) = R [ (1)/( 2 ^(2)) - (1)/(3 ^(2)) ] = (5R)/(36)`
or `lamda _(1) = (36)/(5R) = (36 )/( 5 xx 1. 097 xx 10 ^(7)) = 6563Å`
This wavelength belongs to the Balmer series and lies in the visible region.
(b) For n =2 to n =1,
`(1)/(lamda _(2)) = R [ (1)/( 1 ^(2)) - (1)/( 2 ^(2)) ] = (3R)/(4)`
or `lamda _(2) = (4)/( 3 R) = (4)/( 3 xx 1. 097 xx 10 ^(7)) = 1215Å`
`lamda_(2)` belongs also belongs to the Lyman series and lies in the ultraviolet region.
(c ) For the direct jump n = 3 to n =1,
`(1)/( lamda _(3)) = R [ (1)/( 1 ^(2)) - (1)/(3 ^(2)) ] = (8 R)/(9)`
or `lamda _(3) = (9)/( 8 R) = (9)/( 8 xx 1. 0 9 7 xx 10 ^(7)) = 1026 Å`
Which also belongs to the Lyman series and lies in the ultraviolet region.