$$\eqalign{ & \frac{{{I_0}}}{I} = 4,x = 7\,mm\left( {{\text{given}}} \right) \cr & \Rightarrow \mu = \frac{{2.303\,{{\log }_{10}}\frac{{{I_0}}}{I}}}{x} \cr & \mu = \frac{{2.303\,{{\log }_{10}}4}}{7} = \frac{{2.303 \times 0.6023}}{7} \cr & \Rightarrow \mu = 0.198\,m{m^{ - 1}} \cr} $$

From question, $${m_A} = M;{m_B} = \frac{m}{2}$$
$${u_A} = V,{u_B} = 0$$
Let after collision velocity of $$A = {V_1}$$  and velocity of $$B = {V_2}$$
Applying law of conservation of momentum,
$$mu = m{v_1} + \left( {\frac{m}{2}} \right){v_2}\,\,{\text{or,}}\,\,2\mu = 2{v_1} + {v_2}\,.......\left( {\text{i}} \right)$$
By law of collision
$$e = \frac{{{v_2} - {v_1}}}{{u - 0}}\,\,{\text{or,}}\,\,u = {v_2} - {v_1}\,.......\left( {{\text{ii}}} \right)\,\,\left[ {\because {\text{collision is elastic,}}\,e = 1} \right]$$
using eqns (i) and (ii)
$${v_1} = \frac{1}{3}\mu \,\,{\text{and}}\,\,{v_2} = \frac{4}{3}\mu $$
de-Broglie wavelength $$\lambda = \frac{h}{p}$$
$$\therefore \frac{{{\lambda _A}}}{{{\lambda _B}}} = \frac{{{P_B}}}{{{P_A}}} = \frac{{\frac{m}{2} \times \frac{4}{3}u}}{{m \times \frac{1}{3}}} = 2$$

The minimum energy required to remove an electron from the surface of metal without giving any kinetic energy is called work function.
$${W_0} = h{\nu _0} = \frac{{hc}}{{{\lambda _0}}}$$
Given wavelength,
$$\eqalign{ & {\lambda _0} = 2000\,\mathop {\text{A}}\limits^ \circ \cr & = 2000 \times {10^{ - 10}}m \cr & \therefore {W_0} = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{2000 \times {{10}^{ - 10}}}} \cr & = 9.9 \times {10^{ - 19}}J \cr & = \frac{{9.9 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}}\,\left[ {1\,eV = 1.6 \times {{10}^{ - 19}}\,J} \right] \cr & = 6.2\,eV \cr} $$

Work function for wavelength of $$4100\,\mathop {\text{A}}\limits^ \circ $$  is given by
$$\eqalign{ & {W_0} = \frac{{hc}}{\lambda } = \frac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{4100 \times {{10}^{ - 10}}}} \cr & = 4.8 \times {10^{ - 19}}J \cr} $$
Energy in $$eV$$ is given by
$$\eqalign{ & = \frac{{4.8 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}}\,eV \cr & = 3\,eV \cr} $$
Now, we have
$$\eqalign{ & {W_A} = 1.92\,eV, \cr & {W_B} = 2.0\,eV, \cr & {W_C} = 5\,eV, \cr} $$
Since, $${W_A} < W$$
and $${W_B} < W,$$  hence, $$A$$ and $$B$$ will emit photoelectrons.

According to Bragg’s law,
$$ \Rightarrow 2d\sin \theta = n\lambda ,$$
$$n = 1$$  for first order
$$\eqalign{ & \Rightarrow 2 \times 2.82\sin 15.8 = \lambda \cr & \Rightarrow \lambda = 5.64 \times 0.2723 = 1.53\,\mathop {\text{A}}\limits^ \circ \cr} $$

Independent of frequency $$\left( \nu \right)$$ of light, it only depends on the intensity of incident light. If intensity increases, number of photo electrons increases.

When a photon of light of frequency $$\nu $$ is incident on a photosensitive metal surface. Then, according to Einstein's photoelectric equation
$$h\nu = {W_0} + \frac{1}{2}m{\nu ^2}$$
where, $${W_0}$$ is work function of the metal and $$\frac{1}{2}m{v^2}$$  is $$KE$$  of the photoelectron.
Given, $${W_0} = 3.5\,eV,\,KE = 1.2\,eV$$
$$\therefore h\nu = 3.5 + 1.2 = 4.7\,eV$$
NOTE
Stopping potential always gives maximum kinetic energy of ejected electrons.

$$\eqalign{ & \lambda = \frac{h}{{\sqrt {2meV} }} = {\left( {\frac{{150}}{V}} \right)^{\frac{1}{2}}}\mathop {\text{A}}\limits^ \circ \cr & = {\left( {\frac{{150}}{{150}}} \right)^{\frac{1}{2}}}\mathop {\text{A}}\limits^ \circ \cr & = 1\mathop {\text{A}}\limits^ \circ \cr} $$

According to Einstein's photoelectric equation, $$KE$$ of the photoelectron
\[\frac{1}{2}m{v^2} = \frac{{hc}}{\lambda } - {W_0}\,\,\left[ {\begin{array}{*{20}{c}} {{\rm{where,}}\,{W_0} = {\rm{work}}\,{\rm{function}}}\\ {\frac{1}{2}m{v^2} = KE\,{\rm{of}}\,{\rm{ejected\,electrons}}}\\ {\frac{{hc}}{\lambda } = {\rm{threshold\,energy}}} \end{array}} \right]\]
$$\eqalign{ & {\text{Given,}}\,\,\lambda = 3000\,\mathop {\text{A}}\limits^ \circ = 3000 \times {10^{ - 10}}m \cr & {W_0} = 1\,eV = 1.6 \times {10^{ - 19}}J \cr & m = 9.1 \times {10^{ - 31}}kg \cr & \therefore \frac{1}{2} \times 9.1 \times {10^{ - 31}} \times {v^2} \cr & = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{3000 \times {{10}^{ - 10}}}} - 1.6 \times {10^{ - 19}} \cr & \Rightarrow v = {10^6}m/s \cr} $$

$$\eqalign{ & F = \frac{{dP}}{{dt}} = \left( {\frac{{IA}}{{\frac{{hc}}{\lambda }}}} \right)\left( {1 + r} \right)\frac{h}{\lambda } \cr & \Rightarrow P = \frac{F}{A} = \frac{I}{c}\left( {1 + r} \right) \cr} $$