If $${v_{\max }}$$ is the speed of the fastest electron emitted from the metal surface, then
$$\eqalign{ & \frac{{hc}}{\lambda } = {W_0} + \frac{1}{2}mv_{\max }^2 \cr & \frac{{\left( {6.63 \times {{10}^{ - 34}}} \right) \times \left( {3 \times {{10}^8}} \right)}}{{\left( {180 \times {{10}^{ - 9}}} \right)}} \cr & = 2 \times \left( {1.6 \times {{10}^{ - 19}}} \right) + \frac{1}{2}\left( {9.1 \times {{10}^{ - 31}}} \right)v_{\max }^2 \cr & \therefore v = 1.31 \times {10^6}\,m/s \cr} $$
The radius of the electron is given by
$$r = \frac{{mv}}{{qB}} = \frac{{\left( {9.1 \times {{10}^{ - 31}}} \right) \times \left( {1.31 \times {{10}^6}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}} \right) \times \left( {5 \times {{10}^{ - 9}}} \right)}} = 0.149\,m$$

Applying conservation of linear momentum,
Initial momentum = Final momentum
$$\eqalign{ & 0 = {m_1}{v_1} - {m_2}{v_2} \Rightarrow {m_1}{v_1} = {m_2}{v_2} \cr & {\text{Now,}}\,\frac{{{\lambda _1}}}{{{\lambda _2}}} = \frac{{\frac{h}{{{m_1}{v_1}}}}}{{\frac{h}{{{m_2}{v_2}}}}} = 1 \cr} $$

According to Planck’s quantum theory, a source of radiation emits energy in the form of photons, which travel in straight line. The energy of a photon is given by
$$E = h\nu = \frac{{hc}}{\lambda }$$
where, $$h =$$  Planck's constant $$ = 6.62 \times {10^{ - 34}}J{\text{ - }}s$$
$${\nu _1}$$ and $$\lambda =$$  frequency and wavelength of photon, respectively
$${\text{and}}\,c = {\text{speed}}\,{\text{of}}\,{\text{light}} = 3 \times {10^8}m/s$$

$${\lambda _{\min }} = \frac{{12400}}{{V}}\,\mathop {\text{A}}\limits^ \circ = \frac{{12400}}{{30000}} = 0.413\,\mathop {\text{A}}\limits^ \circ $$

By work function of a metal, it means that the minimum energy required for the electron in the highest level of conduction band to get out of the metal. The work function has no effect on photoelectric current as long as $$h\nu > {W_0}.$$  The photoelectric current is proportional to the intensity of incident light. Since, there is no change in the intensity of light, hence $${I_1} = {I_2}$$

The radiation energy is given by $$E = \frac{{hc}}{\lambda }$$
Initial momentum of the radiation is
$${P_i} = \frac{h}{\lambda } = \frac{E}{c}$$
The reflected momentum is
$${P_r} = - \frac{h}{\lambda } = - \frac{E}{c}$$
So, the change in momentum of light is
$$\Delta {P_{{\text{light}}}} = {P_r} - {P_i} = - \frac{{2E}}{c}$$
Thus, the momentum transferred to the surface is
$$\Delta {P_{{\text{light}}}} = \frac{{2E}}{c}$$

From the graph it is clear that $$A$$ and $$B$$ have the same stopping potential and therefore, the same frequency. Also, $$B$$ and $$C$$ have the same intensity.

Number of emitted electrons $${N_E} \propto {\text{Intensity}} \propto \frac{1}{{{{\left( {{\text{Distance}}} \right)}^2}}}$$
Therefore, as distance is doubled, $${N_E}$$ decreases by $$\left( {\frac{1}{4}} \right)$$ times.

de-Broglie wavelength,
$$\eqalign{ & \lambda = \frac{h}{p} = \frac{h}{{\sqrt {2.m.\left( {K.E} \right)} }} \cr & \therefore \lambda \propto \frac{1}{{\sqrt {K.E} }} \cr} $$
If $$K.E$$  is doubled, wavelength becomes $$\frac{\lambda }{{\sqrt 2 }}$$

$$\eqalign{ & {\lambda _1} = \frac{{hc}}{{e{v_1}}};{\lambda _2} = \frac{{hc}}{{e{v_2}}} \cr & \Rightarrow \frac{{{\lambda _2}}}{{{\lambda _1}}} = \frac{{{v_1}}}{{{v_2}}} \cr & \Rightarrow \frac{3}{4} = \frac{{{v_1}}}{{{v_2}}} \cr & \Rightarrow \left( {\frac{{{v_2} - {v_1}}}{{{v_1}}}} \right) \times 100 = \frac{{100}}{3} \cr} $$
Hence, $$P.D.$$  must be increased by $$\frac{{100}}{3}\% .$$