For I^st line of Lyman series in hydrogen spectrum
$$\frac{1}{\lambda } = R\left( 1 \right)\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) = \frac{3}{4}R$$
For $$L{i^{2 + }}\left[ {\frac{1}{\lambda } = R \times 9\left( {\frac{1}{{{3^2}}} - \frac{1}{{{6^2}}}} \right) = \frac{{3R}}{4}} \right]$$
which is satisfied by $$n = 6 \to n = 3.$$

Impact parameter is perpendicular distance of the velocity vector of the alpha particle from the central line of the nucleus, when the particle is far away from the nucleus of the atom.
Rutherford calculated analytically, the relation between the impact parameter $$b$$ and scattering angle $$\theta ,$$ which is given by
$$b = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{Z{e^2}\cot \frac{\theta }{2}}}{E}$$
where, $$E = \frac{1}{2}m{v^2}$$   is kinetic energy of alpha particle, when it is far away from the atom. According to problem,
$$b = \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{{Z{e^2}\cot \frac{\theta }{2}}}{E} = 0$$
As given that $$b = 0$$
$$\eqalign{ & {\text{so,}}\,\,\cot \frac{\theta }{2} = 0 \cr & \therefore \frac{\theta }{2} = {90^ \circ }\,\,{\text{or}}\,\,\theta = {180^ \circ } \cr} $$

KEY CONCEPT : We know that radius of the nucleus $$R = {R_0}{A^{\frac{1}{3}}},$$   where $$A$$ is the mass number.
$$\eqalign{ & \therefore {R^3} = R_0^3A \cr & \Rightarrow \frac{4}{3}\pi {R^3} = \frac{4}{3}\pi R_0^3A \Rightarrow {\text{Volume}} \propto {\text{mass}}{\text{.}} \cr} $$

According to Bohr, electron can revolve only in certain discrete non-radiating orbits, called stationary orbits, for which total angular momentum of the revolving electron is an integral multiple of $$\frac{h}{{2\pi }},$$  where $$h$$ is Planck’s constant. For orbits, conservation of angular momentum is applicable.
∴ For any permitted orbit, $$mvr = \frac{{nh}}{{2\pi }}$$

Energy of electron in n^th orbit is
$${E_n} = - \left( {Rch} \right)\frac{{{Z^2}}}{{{n^2}}} = - 54.4\,eV$$
For $$H{e^ + }$$  is ground state
$$\eqalign{ & {E_1} = - \left( {Rch} \right)\frac{{{{\left( 2 \right)}^2}}}{{{{\left( 1 \right)}^2}}} = - 54.4 \cr & \Rightarrow Rch = 13.6 \cr} $$
∴ For $$L{i^{ + + }}$$  in first excited state $$\left( {n = 2} \right)$$
$$E' = - 13.6 \times \frac{{{{\left( 3 \right)}^2}}}{{{{\left( 2 \right)}^2}}} = - 30.6\,eV$$

$$\eqalign{ & \frac{{m{v^2}}}{r} = \frac{{3{q^2}}}{{4\pi {\varepsilon _0}{r^2}}} \cr & \Rightarrow mvr = \frac{{3{q^2}}}{{4\pi {\varepsilon _0}v}}\,......\left( {\text{i}} \right) \cr & {\text{and}}\,\,\frac{{nh}}{{2\pi }} = mvr\,......\left( {{\text{ii}}} \right) \cr} $$
Using (i) and (ii) putting $$n=1$$
$$\eqalign{ & \frac{h}{{2\pi }} = \frac{{3{q^2}}}{{4\pi {\varepsilon _0}v}} \cr & \Rightarrow v = \frac{{3{q^2}}}{{2{\varepsilon _0}h}} \cr} $$

For second excited state, $$n = 3$$
$$\therefore $$ Energy needed to ionise $$H$$-atom from its second excited state
$$E = \frac{{2{\pi ^2}mk{e^4}}}{{{h^2}}}\,\left( {\frac{1}{{{3^2}}} - \frac{1}{\infty }} \right)$$
or we can say that
$$\eqalign{ & E \propto \frac{{{Z^2}}}{{{n^2}}}\,\,\left( {_{n\,\, = \,\,{n^{{\text{th}}}}\,\,{\text{orbit}}}^{Z = \,\,{\text{atomic}}\,{\text{number}}}} \right) \cr & {\text{So,}}\,\,E = \frac{{13.6}}{{{3^2}}}eV = 1.51\,eV \cr} $$

From Coulomb's attraction between the positive proton and negative electron $$ = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{e^2}}}{{{r^2}}}\,\,\left[ {{\text{For neutral atom}}} \right]$$
Centripetal force has magnitude $$F = \frac{{m{v^2}}}{r}$$
So for the revolving electrons
$$\eqalign{ & \frac{{m{v^2}}}{r} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{e^2}}}{{{r^2}}} \cr & \Rightarrow {v^2} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{e^2}}}{{mr}} \cr & {\text{or}}\,\,v = \frac{e}{{\sqrt {4\pi {\varepsilon _0}mr} }} \cr} $$
For ground state of $$H$$-atom, $$r = {a_0}$$
$$\therefore v = \frac{e}{{\sqrt {4\pi {\varepsilon _0}m{a_0}} }}$$

$$\eqalign{ & \frac{1}{\lambda } = R\left[ {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right] \cr & \Rightarrow \frac{1}{{970.6 \times {{10}^{ - 10}}}} = 1.097 \times {10^7}\left[ {\frac{1}{{{1^2}}} - \frac{1}{{n_2^2}}} \right] \cr & \Rightarrow {n_2} = 4 \cr} $$
$$\therefore $$ Number of emission line $$N = \frac{{n\left( {n - 1} \right)}}{2} = \frac{{4 \times 3}}{2} = 6$$

For Lyman series for $$H$$-atom
$$\frac{{hc}}{\lambda } = Rhc\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right)$$
and for $$H$$-like ion,
$$\eqalign{ & \frac{{hc}}{\lambda } = {Z^2}Rhc\left( {\frac{1}{{{2^2}}} - \frac{1}{{{4^2}}}} \right) \cr & \therefore \left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right) = {Z^2}\left( {\frac{1}{4} - \frac{1}{{16}}} \right) \cr & \left( {1 - \frac{1}{4}} \right) = {Z^2}\left( {\frac{1}{4} - \frac{1}{{16}}} \right) \cr & Z = 2 \cr} $$