Cut-off wavelength occurs when incoming electron looses its complete energy in collision. This energy appears in the form of X-rays.
Given, mass of electrons = $$m$$
de-Broglie wavelength = $$\lambda $$
So, kinetic energy of electron $$ = \frac{{{p^2}}}{{2m}}$$
$$ = \frac{{{{\left( {\frac{h}{\lambda }} \right)}^2}}}{{2m}} = \frac{{{h^2}}}{{2m{\lambda ^2}}}$$
Now, maximum energy of photon can be given by
$$\eqalign{ & E = \frac{{hc}}{{{\lambda _0}}} = \frac{{{h^2}}}{{2m{\lambda ^2}}} \cr & \Rightarrow {\lambda _0} = \frac{{hc \times 2{\lambda ^2}.m}}{{{h^2}}} \cr & = \frac{{2mc{\lambda ^2}}}{h} \cr} $$

Einstein's photoelectric equation can be written as
$$\eqalign{ & \left( {KE} \right)\frac{1}{2}m{\nu ^2} = h\nu - {W_0} \cr & {\text{where,}}\,\,\left[ {{W_0} = {\text{work function}}} \right] \cr} $$
When photon of energy $$2\,h\,{v_0}$$  falls then
$$ \Rightarrow \frac{1}{2}m \times {\left( {4 \times {{10}^6}} \right)^2} = 2\,h{\nu _0} - h{\nu _0}\,......\left( {\text{i}} \right)$$
When photon of energy $$5\,h\,{v_0}$$  falls then
$$ \Rightarrow \frac{1}{2}m \times {\nu ^2} = 5\,h{\nu _0} - h{\nu _0}\,......\left( {{\text{ii}}} \right)$$
Dividing Eq. (i) by (ii), we get
$$\eqalign{ & \frac{{{\nu ^2}}}{{{{\left( {4 \times {{10}^6}} \right)}^2}}} = \frac{{4\,h{\nu _0}}}{{h{\nu _0}}} \cr & {\text{or}}\,\,{\nu ^2} = 4 \times 16 \times {10^{12}} \cr & {\text{or}}\,\,{\nu ^2} = 64 \times {10^{12}} \cr & \therefore \nu = 8 \times {10^6}m/s \cr} $$

Thermionic emission is the phenomenon of emission of electrons from the metal surface when heated suitably. Here, the energy required for the emission of electrons from metal surface is being supplied by thermal energy. The emitted electrons are called thermal electrons or thermions.

$$\eqalign{ & hv = \phi + {K_{\max }} \cr & hv = 2 + 4 = 6\,eV \cr & {\text{Now,}}\,h\left( {\frac{5}{3}v} \right) = 2 + {K_{\max }} \cr & {K_{\max }} = \frac{5}{3} \times 6 - 2 \cr & \Rightarrow {K_{\max }} = 8\,eV. \cr} $$

(A) Stopping potential is the negative potential applied to stop the photoelectrons from the metal. Stopping potential is given by $$e{V_0} = {K_{\max }}$$
Since, maximum kinetic energy of photoelectrons does not depend on intensity of light, so stopping potential does not vary with intensity of light. Thus, choice (A) is not correct.
(B) If intensity of incident light is increased, we can say that the number of photons incident per unit area per unit time will be increased. Therefore, more electrons will be emitted per second. Hence, photo-current increases. Thus, choice (B) is correct.
(C) The photo-current does not depend on frequency of light. Thus, choice (C) is not correct.
(D) The photo-current does not depend on applied voltage. Thus, choice (D) is not correct.

de-Broglie wavelength of a charged particle is given by, $$\lambda \propto \frac{1}{{\sqrt {mq} }}$$
If $${m_p}$$ and $$e$$ are mass and charge of a proton respectively,
and, $${m_D}$$ and $$e$$ are mass and charge of a deutron respectively, then
$$\eqalign{ & \frac{{{\lambda _p}}}{{{\lambda _D}}} = \sqrt {\frac{{{m_D}{q_D}}}{{{m_p}{q_p}}}} = \sqrt {\frac{{\left( {2{m_p}} \right)\left( e \right)}}{{\left( {{m_p}} \right)\left( e \right)}}} = \sqrt 2 \cr & {\lambda _p} = \sqrt 2 {\lambda _D} \cr} $$
Momentum, is given by, $$P = \frac{h}{\lambda }\,\,{\text{or,}}\,\,p \propto \frac{1}{\lambda }$$
where, $$h$$ = plank's constant
Since the wavelength of a proton is more than that of deutron thus, the momentum of a proton is lesser than that of deutron. Hence, the momentum of proton is less.

Since in the graph retarding potential is same in graph (a) and (b) and photo current is different so, for curves they have same frequency but different intensity of light.

The minimum wavelength below which no photoelectron can emit from metal surface is called cut-off wavelength or threshold wavelength and is given by
$$\eqalign{ & {\text{Work function}} = \frac{{hc}}{{{\text{cut - off wavelength}}}} \cr & {\text{or}}\,\,{\text{cut - off wavelength}} = \frac{{hc}}{{{\text{work}}\,{\text{function}}}} \cr & \therefore {\lambda _0} = \frac{{hc}}{{{W_0}}}\,.......\left( {\text{i}} \right) \cr & {\text{Given,}}\,h = 6.6 \times {10^{ - 34}}\,J - s \cr & c = 3 \times {10^8}m/s \cr & {W_0} = 4.125\,eV \cr & = 4.125 \times 1.6 \times {10^{ - 19}}\,J \cr} $$
Substituting the given values in Eq. (i), we get
$$\eqalign{ & {\lambda _0} = \frac{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{4.125 \times 1.6 \times {{10}^{ - 19}}}}\,\mathop {\text{A}}\limits^ \circ \cr & = 3 \times {10^{ - 7}}m \cr & = 3000\,\mathop {\text{A}}\limits^ \circ \cr} $$

When a photon of light of frequency $$\nu $$ incident on a photosensitive metal surface, the energy of the photon $$\left( {h\nu } \right)$$   is spent in two ways. A part of energy of photon is used in liberating the electron from the metal surface which is equal to the work function $${W_0}$$ of the metal.
$${W_0} = h{\nu _0}$$
(where, $${{\nu _0}}$$ is threshold frequency)
$$\eqalign{ & {\text{or}}\,\,{W_0} = \frac{{hc}}{{{\lambda _0}}} \cr & {\text{Here,}}\,\,{\lambda _0} = 5000\,\mathop {\text{A}}\limits^ \circ \cr & = 5000 \times {10^{ - 10}}m \cr & \therefore {W_0} = \frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{5000 \times {{10}^{ - 10}}}} \cr & = 4 \times {10^{ - 19}}J \cr} $$
NOTE
Threshold frequency $$\left( {{\nu _0}} \right) \to $$   It is the minimum frequency given to metallic surface so that emission of electrons start.

If energy $$E$$ is expressed in $$\left( {eV} \right)$$  and wavelength $$\lambda $$ (in $$\mathop {\text{A}}\limits^ \circ $$ ), then energy of photon,
$$\eqalign{ & E = \frac{{hC}}{\lambda } = \frac{{12375}}{{\lambda \left( {\mathop {\text{A}}\limits^ \circ } \right)}}eV \cr & \therefore \lambda = \frac{{12375}}{{E\left( {eV} \right)}}\mathop {\text{A}}\limits^ \circ \,\,\left[ {\because hc = 12375\,eV - \mathop {\text{A}}\limits^ \circ } \right] \cr & = \frac{{12375}}{{3\left( {eV} \right)}}\mathop {\text{A}}\limits^ \circ = 4125\,\mathop {\text{A}}\limits^ \circ \cr & = 412.5\,nm \cr} $$
NOTE
Energy of photon is $$E = \frac{{hc}}{{\lambda \left( {\mathop {\text{A}}\limits^ \circ } \right)}} = \frac{{12375}}{{\lambda \left( {\mathop {\text{A}}\limits^ \circ } \right)}}eV$$
Here, $$hc = 12375\,eV - \mathop {\text{A}}\limits^ \circ $$     comes from the following procedure
$$\eqalign{ & hc = \left( {{\text{Planck's}}\,{\text{constant}}} \right)\left( {{\text{velocity}}\,{\text{of}}\,{\text{light}}} \right) \cr & = \frac{{\left( {6.6 \times {{10}^{ - 34}}\;J - s} \right)\left( {3 \times {{10}^8}\;m/s} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}\,J/eV} \right)}} \cr & = 12.375 \times {10^{ - 7}}eV - m \cr & = 12375\,eV - \mathop {\text{A}}\limits^ \circ \cr} $$