The de-Broglie wavelength is given by
$$\lambda = \frac{h}{P} \Rightarrow P\lambda = h$$
This equation is in the form of $$yx = c,$$   which is the equation of a rectangular hyperbola. Hence, the graph given in option (B) is the correct one.

$$\eqalign{ & hv_0^2 - h{v_0} = \frac{1}{2}m{v^2} \cr & \therefore \frac{4}{3}h{v_0} - h{v_0} = \frac{1}{2}m{{v'}^2} \cr & \therefore \frac{{{{v'}^2}}}{{{v^2}}} = \frac{{\frac{4}{3}v - {v_0}}}{{v - {v_0}}} \cr & \therefore v' = v\sqrt {\frac{{\frac{4}{3}v - {v_0}}}{{v - {v_0}}}} \cr & \therefore v' > v\sqrt {\frac{4}{3}} \cr} $$

Momentum of photon = $$\frac{E}{c}$$
Change in momentum = $$\frac{{2E}}{c}$$
= momentum transferred to the surface
(the photon will reflect with same magnitude of momentum in opposite direction)

For photoelectric emission from given metal plate, the incident wavelength must be less than that of ultraviolet rays assuming the wavelength of ultraviolet rays as the threshold value. Out of the given radiations, X-rays have wavelength less than that of ultraviolet rays. Thus, X-rays can cause photoelectric emission.

Intensity of light source is given by
$$I \propto \frac{1}{{{d^2}}}$$
where, $$d$$ is the distance of light source from the cell.
So, for two different situations for intensities,
$$\eqalign{ & {\text{or}}\,\,\frac{{{I_1}}}{{{I_2}}} = {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^2} = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4} \cr & {\text{or}}\,\,{I_2} = 4\,{I_1} \cr} $$
As number of photoelectrons emitted is directly proportional to intensity, so number of photoelectrons emitted will become 4 times, i.e. $$4\,n.$$

$$\eqalign{ & {\lambda _{\min }} = 1\mathop {\text{A}}\limits^ \circ \,\left( {{\text{given}}} \right) \cr & \because {\lambda _{\min }} = \frac{{1240}}{E}\left( {eV} \right)\left( {nm} \right) \cr & {\text{Thus, }}E = \frac{{1240\left( {eV} \right)\left( {nm} \right)}}{{0.01\left( {nm} \right)}} = 12400\,eV \cr & E = 12.4\,KeV \cr} $$

By Einstein’s equation of photo-electric effect, the maximum kinetic energy of emitted photo-electrons is given by
$${E_k} = hv - W\,\,{\text{or}}\,\,{E_k} = \frac{{hc}}{\lambda } - W$$
Where, $$h$$ = Planck’s constant
$$v$$ = frequency of incident light
$$W$$=work function of metal
$$\lambda $$ = wavelength of incident light
$$\eqalign{ & {E_k} = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{2000 \times {{10}^{ - 10}} \times 1.6 \times {{10}^{ - 19}}}}eV - 4.2\,eV \cr & {\text{So,}}\,{E_k} = 2\,eV \cr} $$

In a photoelectric effect, when monochromatic radiations of suitable frequency fall on the photosensitive plate called cathode, the photoelectrons are emitted which get accelerated towards anode. These electrons flow in the outer circuit resulting in the photoelectric current.

Using the incident radiations of a fixed frequency, it is found that the photoelectric current increases linearly with the intensity of incident light as shown in figure. Hence, a photocell employs photoelectric effect to convert change in the intensity of illumination into a change in photoelectric current.

According to Einstein's photoelectric equation, kinetic energy of photoelectron
$$KE = h\nu - {W_0}\,\,{\text{or}}\,\,h\nu = KE + {W_0}$$
As maximum KE of ejected electrons is given by
$$KE = e{V_0}$$
where, $${V_0}$$ is stopping potential.
$$\eqalign{ & h\nu = 5\,eV + 6.2\,eV\,\,\left[ {\because e{V_0} = 5\,eV} \right] \cr & = 11.2\,eV\,\,\left( {{\text{lies in X - ray region}}} \right) \cr} $$

Use Einstein’s photoelectric equation.
$$\eqalign{ & {\text{We know that, }}E = {\left( {KE} \right)_{\max }} + {\text{Work function}}\left( \phi \right) \cr & {\text{where,}}\,\,\phi = h{\nu _0},E = h\nu \cr & {\left( {KE} \right)_{\max }} = \frac{1}{2}m\nu _0^2 \cr & \Rightarrow {\left( {KE} \right)_{\max }} = h\nu - \phi \cr & \Rightarrow 2\,eV = 5\,eV - \phi \,\left( {{\text{given}}} \right) \cr & \Rightarrow \phi = 3\,eV \cr & {\text{Thus,}}\,{V_{{\text{cathode}}}} - {V_{{\text{anode}}}} = 3\,V \cr & \Rightarrow {V_{{\text{anode}}}} - {V_{{\text{cathode}}}} = - 3\,V \cr} $$