Number of spectral lines
$$\eqalign{ & N = \frac{{n\left( {n - 1} \right)}}{2} \Rightarrow \frac{{n\left( {n - 1} \right)}}{2} = 6 \cr & {\text{or}}\,\,{n^2} - n - 12 = 0 \cr & {\text{or}}\,\,\left( {n - 4} \right)\left( {n + 3} \right) = 0 \cr & {\text{or}}\,\,n = 4 \cr} $$
Now as the first line of the series has the maximum wavelength, therefore electron jumps from the fourth orbit to the third orbit.

For 2 to 1, 3 to 2 and 4 to 2 we get energy that $$n =4$$  to $$n = 3,$$
I.R. radiation has less energy than U. V. radiation.

$$\eqalign{ & \frac{{m{v^2}}}{r} = \frac{K}{r}\,\,{\text{or}}\,\,v = \sqrt {\frac{K}{m}} = c{r^ \circ } \cr & {\text{Now}}\,\,mvr = \frac{{nh}}{{2\pi }}\,\,{\text{or}}\,\,r = \frac{{nh}}{{2\pi mv}} \cr & \therefore r \propto n. \cr} $$
Also kinetic energy
$$K = \frac{{m{v^2}}}{2} = \frac{m}{2} \times \frac{K}{m} = \frac{K}{2}.$$

For first line of Balmer series
$$\eqalign{ & \frac{1}{\lambda } = R\left( {\frac{1}{4} - \frac{1}{9}} \right) \cr & \Rightarrow R = \frac{{36}}{{5\lambda }} \cr} $$
$$\therefore $$ Wavelength of the first line, $${\lambda _L}$$ of the Lyman series is given by
$$\eqalign{ & \frac{1}{{{\lambda _L}}} = R\left( {1 - \frac{1}{4}} \right) = \frac{{36}}{{5\lambda }} \times \frac{3}{4} = \frac{{27}}{{5\lambda }} \cr & \Rightarrow {\lambda _L} = \frac{{5\lambda }}{{27}} \cr} $$

Radii of Bohr's stationary orbit is given by
$$r = \frac{{{n^2}{h^2}}}{{4{\pi ^2}mk{e^2}Z}} \Rightarrow r \propto \frac{{{n^2}}}{Z}$$
Considering two situations of electrons,
$$ \Rightarrow \frac{{\left( {{r_f}} \right)}}{{\left( {{r_i}} \right)}} = \frac{{n_f^2}}{{n_i^2}}$$
For ground state $${n_i} = 1$$
$$\eqalign{ & \therefore \frac{{21.2 \times {{10}^{ - 11}}}}{{5.3 \times {{10}^{ - 11}}}} = n_f^2 \cr & {\text{or}}\,\,n_f^2 = 4 \cr & \therefore {n_f} = 2 \cr} $$

$$\eqalign{ & \frac{1}{{{\lambda _{\max }}}} = R\left[ {\frac{1}{{{{\left( 1 \right)}^2}}} - \frac{1}{{{{\left( 2 \right)}^2}}}} \right] \cr & \Rightarrow {\lambda _{\max }} = \frac{4}{{3R}} \approx 1213\,\mathop {\text{A}}\limits^ \circ \cr & {\text{and}}\,\,\frac{1}{{{\lambda _{\min }}}} = R\left[ {\frac{1}{{{{\left( 1 \right)}^2}}} - \frac{1}{\infty }} \right] \cr & \Rightarrow {\lambda _{\min }} = \frac{1}{R} \approx 910\,\mathop {\text{A}}\limits^ \circ \cr} .$$

As $$\alpha $$-particles are doubly ionised helium $$H{e^{ + + }}$$  i.e, Positively charged and nucleus is also positively charged and we know that like charges repel each other.

$$\eqalign{ & l = \frac{{nh}}{{2\pi }},\frac{{\left| E \right| \propto {Z^2}}}{{{n^2}}};n = 3 \cr & \Rightarrow {l_H} = {l_{Li}}\,{\text{and}}\,\left| {{E_H}} \right| < \left| {{E_{Li}}} \right| \cr} $$

Wavelength of spectral lines are given by
$$\frac{1}{\lambda } = {z^2}R\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right)$$
For last line of Balmer series,
$$\eqalign{ & {n_1} = 2\,\,{\text{and}}\,\,{n_2} = \infty \cr & \Rightarrow \frac{1}{{{\lambda _B}}} = {z^2}R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{\infty ^2}}}} \right) = \frac{R}{4}\,\,\left[ {\because z = 1} \right] \cr} $$
Similarly, for last line of Lyman series,
$$\eqalign{ & {n_1} = 1\,\,{\text{and}}\,\,{n_2} = \infty \cr & \Rightarrow \frac{1}{{{\lambda _2}}} = {z^2}R\left( {\frac{1}{{{1^2}}} - \frac{1}{{{\infty ^2}}}} \right) = R \cr & \therefore \frac{{\frac{1}{{{\lambda _B}}}}}{{\frac{1}{{{\lambda _L}}}}} = \frac{{\frac{R}{4}}}{R} = \frac{1}{4} \cr & \Rightarrow \frac{{{\lambda _L}}}{{{\lambda _B}}} = \frac{1}{4} \Rightarrow \frac{{{\lambda _B}}}{{{\lambda _L}}} = 4 \cr} $$

$$\eqalign{ & U = - \frac{{k{e^2}}}{{2{R^2}}},F = - \frac{{dU}}{{dR}} = \frac{{3k{e^2}}}{{2{R^4}}} \cr & {\text{But,}}\,F = \frac{{m{v^2}}}{R} \Rightarrow \frac{{m{v^2}}}{R} = \frac{{3k{e^2}}}{{2{R^4}}} \cr & {\text{Also,}}\,\,mvR = \frac{{nh}}{{2\pi }} \cr} $$
Solve to get : $$R = \frac{{6{\pi ^2}k{e^2}m}}{{{n^2}{h^2}}}$$