Electrons of energy `12.09 eV` can excite hydrogen atoms . To which orbit is the electron in the hydrogen atom raisd and what are the wavelengths of the radiations emitted as it dropes back to the ground state?

Explanation

The energy of the electron in different states are:
`E_(1) = - 13.6 e V` for `n = 1`.
`E_(2) = - 3.4 e V` for `n = 2`.
and `E_(3) = - 1.51 e V` 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 e V`. Thus , the electron ia reised to the third orbit of principal quantum number `n = 3` .
Now , an electron in the `n = 3` level can return to the ground state making the following possible jumps:
a. `a = 3` to `n = 2` and then from `n = 2 to n = 1`
b. `n = 3 to n = 1`
Thus, the corresponding wevelength emitting are a.
For `n = 3` to `n = 2`:
`(1)/(lambda_(1)) = R[(1)/(2^(2)) - (1)/(3^(2))] = (5 R)/(36)`
or `lambda_(1) = (36)/(5 R) = (36)/(5 xx 1.097 xx 10^(7)) = 6563 Å`
This wevelengths belongs to the Balmar series and lines in the visible region.
b. For `n = 2 to n = 1:`
`(1)/(lambda_(2)) = R[(1)/(1^(2)) - (1)/(2^(2))] = (3 R)/(4)`
or `lambda_(1) = (4)/(4 R) = (4)/(3 xx 1.097 xx 10^(7)) = 1215 Å`
`lambda_(2)` belongs to the Lyman series and lines in the ultraviolect region.
c. For the direct jump `n = 3 to n = 1`:
`(1)/(lambda_(3)) = R[(1)/(1^(2)) - (1)/(3^(2))] = (8 R)/(9)`
or `lambda_(3) = (9)/(8 R) = (9)/(8 xx 1.097 xx 10^(7)) = 1026 Å`
which also belongs to the Lyman series and lies in the ultraviolet.