As electron of charge $$'e’$$ is passed through $$'V'$$ volt, kinetic energy of electron will be $$eV$$
Wavelength of electron wave $$\left( \lambda \right) = \frac{h}{{\sqrt {2m.K.E} }}$$
$$\lambda = \frac{h}{{\sqrt {2meV} }} \Rightarrow \,\frac{h}{\lambda } = \sqrt {2meV} $$

For electrons present in $$M$$  shell the value of other quantum numbers are same. But, the value of spin quantum number will be different.

(i) 4$$p,$$ (ii) 4$$s,$$ (iii) 3$$d,$$ (iv) 3$$p,$$
The order of increasing energy
(iv) < (ii) < (iii) < (i)

Energy of the orbit increases as we move away from the nucleus or as the value of $$n$$  increases.

$$\eqalign{ & {\text{For}}\,\,n = 3,\,l = 2\,{\text{the subshell is}}\,\,3d\left( {n + l = 5} \right) \cr & n = 4,l = 2\,{\text{the subshell is}}\,\,4d\left( {n + l = 6} \right) \cr & n = 4,\,l = 1\,{\text{the subshell is}}\,\,4p\left( {n + l = 5} \right) \cr & n = 5,\,l = 0,\,{\text{the subshell is}}\,\,5s\left( {n + l = 5} \right) \cr} $$
According to $$(n + l)$$  rule greater the $$(n + l)$$  value, greater the energy that is 6.

$$\eqalign{ & I.E = \frac{{{Z^2}}}{{{n^2}}} \times 13.6\,eV\,...\left( {\text{i}} \right) \cr & {\text{or}}\,\frac{{{I_1}}}{{{I_2}}} = \frac{{Z_1^2}}{{n_1^2}} \times \frac{{n_2^2}}{{Z_2^2}}\,...\left( {{\text{ii}}} \right) \cr & {\text{Given,}} \cr & {I_1} = - 19.6 \times {10^{ - 18}},\, \cr & {Z_1} = 2,{n_1} = 1,{Z_2} = 3\,\,{\text{and}}\,\,{n_2} = 1 \cr & {\text{Substituting these values in equation (ii)}}{\text{.}} \cr & - \frac{{19.6 \times {{10}^{ - 18}}}}{{{I_2}}} = \frac{4}{1} \times \frac{1}{9} \cr & {\text{or}}\,{I_2} = - 19.6 \times {10^{ - 18}} \times \frac{9}{4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = - 4.41 \times {10^{ - 17}}J/atom \cr} $$

In Balmer series, $${n_1} = 2$$   and $${n_2} = 3,4,5$$   So, third line arises by the emission of 5^th orbit to 2^nd orbit.

$$\eqalign{ & \lambda = \frac{h}{{{m_e}x}} \cr & = \frac{h}{{{m_p}V}} \cr & = \frac{h}{{1840\,{m_e}V}}\,\,\left[ {\therefore \,\,{m_p} = 1840\,{m_e}} \right] \cr & {\text{Hence,}}\,\,V = \frac{x}{{1840}} \cr} $$

$$\eqalign{ & {\text{As we know, }} \cr & \bar v = - {R_H}\left( {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right)\,{Z^2}\,\left( {{\text{where}},\,Z = 1} \right) \cr & {\text{After putting the values, we get }} \cr & \bar v = - {R_H}\left( {\frac{1}{{{n^2}}} - \frac{1}{{{8^2}}}} \right) \cr & \therefore \,\,\bar v = \frac{{{R_H}}}{{64}} - \frac{{{R_H}}}{{{n^2}}} \cr & {\text{Comparing to y = }}mx{\text{ + }}c{\text{, we get}} \cr & x{\text{ = }}\,{\text{and}}\,m{\text{ = }}\, - {{\text{R}}_H}\left( {{\text{slope}}} \right) \cr} $$

$$\eqalign{ & {v_n} = {r_n}\omega \,\,{\text{where}}\,\,{r_n} = \frac{{{n^2}{h^2}}}{{4{\pi ^2}m{e^2}Z \cdot K}} \cr & {\text{and}}\,\,{v_n} = \frac{{2\pi \cdot Z \cdot {e^2} \cdot K}}{{n \cdot h}} \cr & \therefore \,\,\frac{{2\pi Z{e^2} \cdot K}}{{n \cdot h}} = \frac{{{n^2}{h^2}}}{{4{\pi ^2}m{e^2}\,Z \cdot K}} \times \omega \,; \cr & \omega = \frac{{8{\pi ^3}m{e^4} \cdot {Z^2} \cdot {K^2}}}{{{n^3} \cdot {h^3}}} \cr & = \frac{{9{\pi ^3}m{e^4} \cdot {K^2}}}{{{h^3}}}\left( {\because \,\,n = 2\,{\text{and}}\,Z = 3\,} \right) \cr} $$