Question

In a photoelectric experiment, a parallel beam of monochromatic light with a 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 500V is applied between the cathode and anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F = n×10-4 N due to the impact of the electrons. The value of n is ______.
Mass of the electron, ${m_e} = 9 \times {10^{ - 31}}kg$and $1eV = 1.6 \times {10^{19}}J$.

Answer 1

From Einstein’s Photoelectric Equation,
$hf = {W_0} + {E_{\max }}$
Where W0= Work function of cathode metal
hf= incident radiation of photons
Emax= maximum kinetic energy attained by emitted electron
We know, $KE = \dfrac{{{{(momentum)}^2}}}{{2 \times mass}} = \dfrac{{{P^2}}}{{2m}}$ and change in energy due to potential difference $\Delta V = q\Delta V$

Complete step by step answer:
The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy,
so, Emax= 0.
Work function, ${W_0} = 6.25eV$.
Now, from Einstein’s equation, $hf = {W_0} + {E_{\max }}$
Now, $hf = 6.25eV = 6.25 \times 1.6 \times {10^{ - 19}}J = {10^{ - 18}}J$.
Power of monochromatic light, W= 200W
Let, N be the no. of photons per second.
Now, $W = Nhf$
Or, $200 = N \times {10^{ - 18}}$
$\therefore N = 2 \times {10^{20}}$
As efficiency is 100%, No. of photons per second= no. of electrons emitted per second.
Let, P be the momentum of one emitted electron and As photon is just above threshold frequency, $KE_max$ is zero and they are accelerated by potential difference of 500V.
Change in energy due to potential difference$\Delta V = q\Delta V$= Change in Kinetic Energy
Now, $q\Delta V = \dfrac{{{P^2}}}{{2m}}$
Or, $P = \sqrt {2mq\Delta V}$
As photons are completely absorbed, force exerted in one second
$= nP \\\ = 2 \times {10^{20}} \times \sqrt {2 \times (9 \times {{10}^{ - 31}}) \times (1.6 \times {{10}^{ - 19}}) \times 500} \\\ = 24 \times {10^{ - 4}}N \\\$
So, the value of n= 24.

The correct answer is 24.

Note:
We can take, no. of photons per second= no. of electrons emitted per second if and only if when the efficiency is 100%. If efficiency is not mentioned we can’t take this assumption.