$${{v'}_1} = \frac{{\left( {{m_1} - {m_2}} \right){v_1} + 2{m_2}{v_2}}}{{{m_1} + {m_2}}}$$
As $${v_2}$$ is zero, $${m_2} > {m_1},{{v'}_1}$$   is in the opposite direction.
$$\eqalign{ & {m_1} = 1,{m_2} = A. \cr & \therefore \left| {{{v'}_1}} \right| = \frac{{\left( {A - 1} \right)}}{{\left( {A + 1} \right)}}{v_1} \cr} $$
The fraction of total energy retained is
$$\frac{{\frac{1}{2}mv'\,_1^2}}{{\frac{1}{2}v_1^2}} = \frac{{{{\left( {A - 1} \right)}^2}}}{{{{\left( {A + 1} \right)}^2}}}$$

For the third line of Balmer series,
$$\eqalign{ & {n_1} = 2,{n_2} = 5 \cr & \therefore \frac{1}{\lambda } = R{Z^2}\left( {\frac{1}{{n_1^2}} - \frac{1}{{n_2^2}}} \right) \cr & = R{Z^2}\left( {\frac{1}{{{2^2}}} - \frac{1}{{{5^2}}}} \right) = \frac{{21R{Z^2}}}{{100}} \cr & E = - 13.6\,eV \cr & {Z^2} \times \frac{{21}}{{100}} = \frac{{hc}}{\lambda } = \frac{{1242\,eVnm}}{{108.5\,nm}} \cr & {Z^2} = \frac{{1242 \times 100}}{{108.5 \times 21 \times 13.6}} = 4 \cr & \Rightarrow Z = 2 \cr} $$

$${E_p} = \frac{{2{K_{{\text{ex}}}}}}{{{{\left( {{x^2} + \frac{{{b^2}}}{4}} \right)}^{\frac{3}{2}}}}}$$
For maximum $${E_P}$$
$$\eqalign{ & \frac{{d{E_P}}}{{dx}} = 0 \cr & \Rightarrow x = \frac{b}{{2\sqrt 2 }} \cr} $$

Speed of electron in first orbit $$\left( {n = 1} \right)$$  of hydrogen atom $$\left( {z = 1} \right),$$
$$\eqalign{ & v = \frac{{{e^2}}}{{2{\varepsilon _0}h}} \cr & r = \frac{{{h^2}{\varepsilon _0}}}{{\pi m{e^2}}} \Rightarrow {\varepsilon _0} = \frac{{r\pi m{e^2}}}{{{h^2}}} \cr} $$
Acceleration of electron,
$$\eqalign{ & \frac{{{v^2}}}{r} = \frac{{{e^4}}}{{4\varepsilon _0^2{h^2}}} \times \frac{{\pi m{e^2}}}{{{h^2}{\varepsilon _0}}} \cr & = \frac{{{e^4} \times \pi m{e^2}}}{{4{h^4}\varepsilon _0^2}}\,......\left( {{\text{ii}}} \right) \cr} $$
eliminating $${\varepsilon _0}$$
$$ = \frac{{{e^4}\pi m{e^2}{h^6}}}{{4{h^4}{r^3}{\pi ^3}{m^3}{e^6}}} = \frac{{{h^2}}}{{4{\pi ^2}{m^2}{r^3}}}$$

$$\eqalign{ & \frac{1}{{{\lambda _1}}} = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right) = \frac{{5R}}{{36}} \cr & \frac{1}{{{\lambda _2}}} = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{4^2}}}} \right) = \frac{{3R}}{{16}} \cr & \therefore \frac{{{\lambda _2}}}{{{\lambda _1}}} = \frac{{80}}{{108}} \cr & {\lambda _2} = \frac{{80}}{{108}}{\lambda _1} = \frac{{80}}{{108}} \times 660 = 488.9nm \cr} $$

To obtain 6 spectral line, as electron must be excited to fourth orbit with energy $$\frac{{{E_g}}}{{16}},$$  so that the difference is $$\frac{{15\,{E_g}}}{{16}}.$$

The energy in the ground state
$$\eqalign{ & {E_1} = - 4\,Rydberg \cr & {E_1} = - 4 \times 2.2 \times {10^{ - 18}}J \cr} $$
The energy of the first excited state $$\left( {n = 2} \right)$$
$${E_2} = \frac{{{E_1}}}{4} = - 2.2 \times {10^{ - 18}}J$$
The energy difference
$$\Delta E = {E_2} - {E_1} = 3 \times 2.2 \times {10^{ - 18}}J$$
Now, the wavelength of radiation emitted is
$$\eqalign{ & \lambda = \frac{{hc}}{{\Delta E}} \cr & \lambda = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3 \times 2.2 \times {{10}^{ - 18}}}} = 300\,\mathop {\text{A}}\limits^ \circ \cr} $$

For $$H$$ atom
$${E_2} - {E_1} = 10.2\,eV$$
For $$Z= 10,$$
$$\eqalign{ & {E_1} = \frac{{ - 13.6 \times {{10}^2}}}{{{1^2}}} = - 1360\,eV \cr & {E_2} = \frac{{ - 13.6 \times {{10}^2}}}{{{2^2}}} = - 3.4 \times 100 = - 340\,eV \cr & \therefore {E_2} - {E_1} = - 340 - \left( { - 1360} \right) = 1020\,eV \cr} $$

A particle of mass $${M_1}$$ and charge $${Z_1}$$ possess initial velocity $$u,$$ when it is at a large distance from the nucleus of an atom having atomic number $${Z_2}.$$  At the distance of closest approach, the kinetic energy of particle is completely converted to potential energy. Mathematically,
$$\frac{1}{2}{M_1}{u^2} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{Z_1}{Z_2}}}{{{r_0}}}$$
So, the energy of the particle is directly proportional to $${Z_1}{Z_2}.$$

When one $${e^ - }$$ is removed from neutral helium atom, it becomes a one $${e^ - }$$ species.
For one $${e^ - }$$ species we know
$${E_n} = \frac{{ - 13.6{Z^2}}}{{{n^2}}}eV/atom$$
For helium ion, $$Z = 2$$ and for first orbit $$n = 1.$$
$$\therefore {E_1} = \frac{{ - 13.6}}{{{{\left( 1 \right)}^2}}} \times {2^2} = - 54.4eV$$
$$\therefore $$ Energy required to remove this $${e^ - } = + 54.4\,eV$$
$$\therefore $$ Total energy required $$ = 54.4 + 24.6 = 79eV$$