Question 35: Dual Nature of Matter and Radiation

Question

In a photoelectric experiment a parallel beam of monochromatic light with power of $$200$$ $$W$$ is incident on a perfectly absorbing cathode of work function $$6.25$$ $$ev.$$ The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron  emission efficiency  is $$100\% $$. A potential difference of $$500$$ $$V$$  is applied between the cathode and the anode. All the emitted electrons are  incident normally on the anode and are absorbed.  The anode experiences a force $$F = n 	imes {10^{ - 4}}$$ $$N$$ due to the impact of the electrons. The value of $$n$$  is ______________. Mass of the electron $${M_e} = 9 	imes {10^{ - 31}}\,kg$$  and $$1.0eV = 1.6 	imes {10^{ - 19}}\,J.$$

Accepted Answer

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
$${F_a} = {N_e}\Delta {p_a} = {N_e}{p_e}$$ ($$\because$$ $$\Delta {p_a} = {p_e}$$)
$$ = {N_p}{p_e} = (P/\phi ){p_e}$$ ($$\because$$ $${N_e} = {N_p} = P/\phi $$)
$$ = (P/\phi )\sqrt {2{m_e}{K_e}} $$ ($$\because$$ $${p_e} = \sqrt {2{m_e}{K_e}} $$)
$$ = (P/\phi )\sqrt {2{m_e}eV} $$ ($$\because$$ $${K_e} = eV$$
$$ = 2.4 	imes {10^{ - 4}}N$$.

Dual Nature of Matter and Radiation

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