Electrons of energy 12.09 eV can excite hydrogen atoms. To which orbit is the electron in the hydrogen atom raised and what are the wavelengths 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 eV " for "n = 1 `
`E_2 = -3.4 eV " for " n = 2`
and `E_3 =-1.51 eV" for "n = 3 `
Evidently, the energy needed 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 electron is raised to the third orbit of principal quantum number n = 3.
Now an electron in the n=3 level can return to the ground state by making the following possible jumps:
(i) n = 3 to n = 2 and then from n = 2 to n = 1.
(ii) 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)/(5xx1.097xx10^7)=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/(3R) = 4/(3xx1.097xx10^7)=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] =(8R)/9`
or `lamda_3=9/(8R)=9/(8xx1.097xx10^(7)) =1026Å`
which also belongs to the Lyman series and lies in the ultraviolet region.