QJEEAdv18Phd04f: JEE Advanced 2018 Physics - Class 12 JEEAdv18Phd04f Solution: In a photoelectric experiment a parallel beam of monochromatic…

The frequency of incident light is just above the threshold frequency. Hence, the energy of each photon is equal to the work function (E_p = $\phi$ = 6.25 eV) and the kinetic energy of emitted photo-electron is zero (K_e0 = 0). The energy incident per second on the cathode is incident power P = 200 W. Thus, number of photons incident per second is

N_p = P/E_p = P/ $\phi$ .

The photo-electron emission efficiency is 100%. Thus, number of photo-electron emitted per second is equal to the number of photons incident per second i.e., N_e = N_p. These photo-electrons are accelerated by a potential difference V = 500 V. Thus, gain in potential energy of each photo-electron is $\Delta$ U = eV = 500 eV. By conservation of energy, kinetic energy of the photo-electron when it reaches the anode is

K_e = K_e0 + $\Delta$ U = 500 eV.

The linear momentum of photo-electron of mass m_e and kinetic energy K_e is given by

${p_e} = \sqrt {2{m_e}{K_e}}$ .

The photo-electron transfer its entire linear momentum to the anode (absorbed by the anode). Thus, gain in linear momentum of the anode by absorbing one photo-electron is

$\Delta$ p_a = p_e.

The force on the anode is equal to the increase in its linear momentum per second. Since N_e photo-electrons strikes the anode per second, the force acting on the anode is given by

${F_a} = {N_e}\Delta {p_a} = {N_e}{p_e}$ ( $\because$ $\Delta {p_a} = {p_e}$ )

$= {N_p}{p_e} = (P/\phi ){p_e}$ ( $\because$ ${N_e} = {N_p} = P/\phi$ )

$= (P/\phi )\sqrt {2{m_e}{K_e}}$ ( $\because$ ${p_e} = \sqrt {2{m_e}{K_e}}$ )

$= (P/\phi )\sqrt {2{m_e}eV}$ ( $\because$ ${K_e} = eV$

$= 2.4 \times {10^{ - 4}}N$ .