Question

If radiation corresponds to second line of ‘Balmer series’ of $ {\text{L}}{{\text{i}}^{ + 2}} $ ion, knocked out electron from first excited state of H-atom, then kinetic energy of ejected electron would be:
(A) $ 2.55{\text{ eV}} $
(B) $ 4.25{\text{ eV}} $
(C) $ 11.25{\text{ eV}} $
(D) $ 19.55{\text{ eV}} $

Answer 1

Hint: We need to calculate the energy difference between the first excited state and infinite excited state. This will give us the value of energy of the ejected electron. The Balmer series starts from n value 2 and the second line goes to 4.

Formula used: $ {\text{E}} = - 13.6{{\text{Z}}^2}\left[ {\dfrac{1}{{{\text{n}}_1^2}} - \dfrac{1}{{{\text{n}}_2^2}}} \right] $
Here E is the energy, Z is nuclear charge, $ {{\text{n}}_1} $ is initial energy state and $ {{\text{n}}_2} $ is final energy state.

Complete step by step solution:
In the Hydrogen spectrum Balmer series is the second set of lines. The ground state energy state starts from n value 2 and goes till infinity. If we will find the energy of the first state and the infinite excited state and subtract both of these, we will get the energy possessed by the electron that is ejected. These energies will be calculated using the Rydberg formula:
 $ {\text{L}}{{\text{i}}^{ + 2}} $ is a hydrogen like species because it also has one electron just like hydrogen so the rydberg formula and energy formula will be applicable here too. The Z value will be 3 here because the nuclear charge on lithium is 3 due to the presence of 3 protons.
The energy difference between ground state and first excited state of second line that is from 2 to 4 will be:
 $ {\text{E}} = - 13.6 \times {{\text{3}}^2}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right] $
 $ \Rightarrow {\text{E}} = 13.6 \times \dfrac{{27}}{{16}} $
Now the energy between ground state and infinity for hydrogen, because electron is ejected from hydrogen atom, will be:
 $ {\text{E}} = - 13.6 \times {1^2}\left[ {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{\infty ^2}}}} \right] = \dfrac{{13.6}}{4} $
The difference in energy will be:
 $ 13.6 \times \dfrac{{27}}{{16}} - \dfrac{{13.6}}{4} = 19.55{\text{ eV}} $
Hence, the correct option is D.

Note:
Hydrogen spectra contain five lines named as layman, balmer, paschen, bracket and pfund. The lowest energy level for hydrogen starts from 1 and not from zero, because if it starts from zero the zero point energy will be zero and this is not possible.