Energy of a photon of frequency $$\nu $$ is given by $$E = h\nu .$$
$$\eqalign{ & {\text{Also,}}\,E = m{c^2},m{c^2} = h\nu \cr & \Rightarrow mc = \frac{{h\nu }}{C} \Rightarrow p = \frac{{h\nu }}{c} \cr} $$

If $$n = $$ no. of channels
$$\eqalign{ & 10\% \,{\text{of}}\,10\,GHz = n \times 5\,KHz \cr & {\text{or,}}\,\frac{{10}}{{100}} \times 10 \times {10^9} = n \times 5 \times {10^3} \Rightarrow n = 2 \times {10^5} \cr} $$

$$\eqalign{ & \frac{{hc}}{{e{\lambda _1}}} - \frac{\phi }{e} = {V_0}_{_1}\,{\text{and}}\,\frac{{hc}}{{e{\lambda _2}}} - \frac{\phi }{e} = {V_0}_{_2} \cr & \therefore \frac{{hc}}{e}\left[ {\frac{1}{{{\lambda _1}}} - \frac{1}{{{\lambda _2}}}} \right] = {V_0}_{_1} - {V_0}_{_2} \cr & \therefore h = \frac{{e\left( {{V_0}_{_1} - {V_0}_{_2}} \right){\lambda _1}{\lambda _2}}}{{\left( {{\lambda _2} - {\lambda _1}} \right)c}} \cr} $$
From the first two values given in data
$$\eqalign{ & h = \frac{{1.6 \times {{10}^{ - 19}}\left[ {2 - 1} \right] \times 0.4 \times 0.3 \times {{10}^{ - 6}}}}{{0.1 \times 3 \times {{10}^8}}} \cr & h = 0.64 \times {10^{ - 33}} = 6.4 \times {10^{ - 34}}J - s \cr} $$
Similarly if we calculate $$h$$ for the last two values of data $$h = 6.4 \times {10^{ - 34}}J - s$$

$$\eqalign{ & \lambda = 400nm,hc = 1240\,eVnm,K.E. = 1.68\,eV \cr & {\text{We}}\,{\text{know}}\,{\text{that}}\,\frac{{hc}}{\lambda } - W = K.E \Rightarrow W = \frac{{hc}}{\lambda } - K.E \cr & \Rightarrow W = \frac{{1240}}{{400}} - 1.68 = 3.1 - 1.68 = 1.42\,eV \approx 1.41\,eV \cr} $$

$${E_0} = C{B_0}{\text{ and }}C = \frac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$$
Electric energy density $$ = \frac{1}{2}{\varepsilon _0}E_0^2 = {\mu _E}$$
Magnetic energy density $$ = \frac{1}{2}\frac{{B{o^2}}}{{{\mu _0}}} = {\mu _B}$$
Thus, $${\mu _E} = {\mu _B}$$
Energy is equally divided between electric and magnetic field

$$\eqalign{ & h\nu _0^2 - h{\nu _0} = \frac{1}{2}m{\nu ^2} \cr & \therefore \frac{4}{3}h{\nu _0} - h{\nu _0} = \frac{1}{2}m\nu {'^2} \cr & \therefore \frac{{\nu {'^2}}}{{{\nu ^2}}} = \frac{{\frac{4}{3}\nu - {\nu _0}}}{{\nu - {\nu _0}}}\,\,\,\,\,\,\therefore \nu ' = \nu \sqrt {\frac{{\frac{4}{3}\nu - {\nu _0}}}{{\nu - {\nu _0}}}} \cr & \therefore \nu ' > \root \nu \of {\frac{4}{3}} \cr} $$

As $$\lambda $$ is increased, there will be a value of $$\lambda $$ above which photoelectrons will be cease to come out so photocurrent will become zero. Hence (D) is correct answer.

For $${K_\alpha },\frac{1}{\lambda } \propto {\left( {Z - 1} \right)^2}$$
From (i), $$\frac{{{\lambda _2}}}{{{\lambda _1}}} - \frac{{{{\left( {{Z_1} - 1} \right)}^2}}}{{{{\left( {{Z_2} - 1} \right)}^2}}} \Rightarrow \frac{{4\lambda }}{\lambda } = \frac{{{{\left( {11 - 1} \right)}^2}}}{{{{\left( {{Z_2} - 1} \right)}^2}}}$$
$$ \Rightarrow {Z_2} - 1 = \frac{{10}}{2} \Rightarrow {Z_2} = 6$$

For the longest wavelength to emit photo electron
$$\eqalign{ & \frac{{hc}}{\lambda } = \phi \Rightarrow \lambda = \frac{{hc}}{\phi } \cr & \Rightarrow \lambda = \frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{40 \times 1.6 \times {{10}^{ - 16}}}} = 310\,nm \cr} $$

By Einstein photoelectric equation,
$${K_{\max }} = e{V_0} = hv - h{v_0}$$
When $$v$$ is doubled, $${K_{\max }}$$  and $${V_0}$$ become more than double.