Energy levels of $$H$$-atom are given by
$$\eqalign{ & {E_n} = - \frac{{13.6{Z^2}}}{{{n^2}}}eV\,\,\left( {Z = 1} \right) \cr & \Rightarrow {E_n} = \frac{{ - 13.6}}{{{n^2}}}\,eV \cr} $$
Photons are emitted only when electron jumps from higher energy level (higher $$n$$-value) to lower energy level (lower $$n$$-value). So, alternative (a) and (c) are wrong.
Energy difference from $$n = 2$$  to $$n = 1$$  level is
$$\Delta {E_{2 \to 1}} = 13.6\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right)eV = 13.6 \times \frac{3}{4} = 10.2\,eV$$
$$\eqalign{ & \Delta {E_{6 \to 2}} = 13.6\left( {\frac{1}{{{2^2}}} - \frac{1}{{{6^2}}}} \right) \cr & = 13.6 \times \left( {\frac{1}{4} - \frac{1}{{36}}} \right) = 13.6 \times \frac{2}{9} = 3.02\,eV \cr} $$
Thus, it is evident that difference is larger for $$n = 2$$  to $$n = 1$$  transition. Hence, maximum energy photon or shortest wavelength will be emitted during transition from $$n =2$$  to $$n = 1$$

KEY CONCEPT : For an atom following Bohr’s model, the radius is given by
$${r_m} = \frac{{{r_0}{m^2}}}{Z}$$   where $${r_0}$$ = Bohr’s radius and $$m$$ = orbit number.
For $$Fm,m = 5$$   (Fifth orbit in which the outermost electron is present)
$$\therefore {r_m} = \frac{{{r_0}{5^2}}}{{100}} = n{r_0}\left( {{\text{given}}} \right) \Rightarrow n = \frac{1}{4}$$

$$\eqalign{ & h{\nu _L} = {E_\infty } - {E_1}\,......\left( {\text{i}} \right) \cr & h{\nu _f} = {E_\infty } - {E_5}\,......\left( {{\text{ii}}} \right) \cr & E\,\infty \,\frac{{{z^2}}}{{{n^2}}} \Rightarrow \frac{{{E_5}}}{{{E_1}}} = {\left( {\frac{1}{5}} \right)^2} = \frac{1}{{25}} \cr & {\text{Equation}}\,\frac{{\left( {\text{i}} \right)}}{{\left( {{\text{ii}}} \right)}} \Rightarrow \frac{{h{\nu _L}}}{{h{\nu _f}}} = \frac{{{E_1}}}{{{E_5}}} \cr & \Rightarrow \frac{{{\nu _L}}}{{{\nu _f}}} = \frac{{25}}{1} \Rightarrow {\nu _f} = \frac{{{\nu _L}}}{{25}} \cr} $$

The energy of hydrogen like atom in its $$n$$^th excited state is given by
$${E_n} = - 13.6\frac{{{Z^2}}}{{{n^2}}}.$$
For ground state $$\left( {n = 1} \right),$$  and atomic number $$\left( Z \right) = 1$$
$${E_1} = - \frac{{13.6}}{{{{\left( 1 \right)}^2}}} = - 13.6\,eV$$
For first excited state $$\left( {n = 2} \right),$$
$$\eqalign{ & {E_2} = - \frac{{13.6}}{{{{\left( 2 \right)}^2}}} = - \frac{{13.6}}{4} \cr & = - 3.4\,eV \cr} $$
NOTE
In ground state $$\left( {n = 1} \right)$$  energy of atom is $$-13.6\,eV$$  and energy corresponding to $$n = \infty $$  is zero. Therefore, energy required to remove the electron from ground state is $$13.6\,eV.$$

In the ground state, $$n = 1$$
$${E_1} = - \frac{{13.6}}{{{1^2}}} = - 13.6\,eV$$
For the first excited state (i.e. for $$n = 2$$ )
$$\eqalign{ & {E_2} = \frac{{ - 13.6}}{{{2^2}}} = - 3.4\,eV \cr & \Delta E = {E_2} - {E_1} \cr & = - 3.4 + 13.6 \cr & = 10.2\,eV \cr} $$

Electron after absorbing $$10.2\,eV$$  energy goes to its first excited state $$\left( {n = 2} \right)$$  from ground state $$\left( {n = 1} \right).$$
$$\therefore $$ Increase in momentum $$ = \frac{h}{{2\pi }}$$
$$ = \frac{{6.6 \times {{10}^{ - 34}}}}{{6.28}} = 1.05 \times {10^{ - 34}}J - s.$$

For helium ion, $$Z=2$$  and for first orbit $$n = 1.$$
$$\therefore {E_1} = \frac{{ - 13.6}}{{{{\left( 1 \right)}^2}}} \times {2^2} = - 54.4\,eV$$
$$\therefore $$ Energy required to remove this $${e^ - } = + 54.4\,eV$$
$$\therefore $$ Total energy required $$= 54.4 + 24.6 =79\,eV$$

$$\eqalign{ & U = - K\frac{{z{e^2}}}{r};T.E = - \frac{k}{2}\frac{{z{e^2}}}{r} \cr & K.E = \frac{k}{2}\frac{{z{e^2}}}{r}.\,{\text{Here}}\,r\,{\text{decreases}} \cr} $$

$$\eqalign{ & I = \frac{e}{T} = \frac{{eV}}{{2\pi r}} \cr & {\text{so,}}\,M = \frac{{eV}}{{2\pi r}} \times \pi {r^2} = \frac{{evr}}{2} \cr} $$
According to Bohr’s theory angular momentum
$$\eqalign{ & mvr = \frac{{nh}}{{2\pi }}\,\,{\text{or}}\,\,vr = \frac{{nh}}{{2\pi m}} \cr & {\text{so,}}\,\,M = \frac{{neh}}{{4\pi m}} \cr} $$
For the ground state $$n = 1\,{\text{so,}}\,\,M = \frac{{eh}}{{4\pi m}}$$

The energy level diagram of the atom is shown in the figure. It is clear that

$$\eqalign{ & {E_{{K_\beta }}} = {E_{{K_\alpha }}} + {E_{{L_\alpha }}} \cr & {\text{or,}}\,\,{v_{{K_\beta }}} = {v_{{K_\alpha }}} + {v_{{L_\alpha }}} \cr & {\text{or,}}\,\,\frac{1}{{{\lambda _{{K_\beta }}}}} = \frac{1}{{{\lambda _{{K_\alpha }}}}} + \frac{1}{{{\lambda _{{L_\alpha }}}}} \cr} $$