Question

In a photoelectric experiment, a parallel beam of monochromatic light with the power of $200 \, \, \text{W}$ 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 $500 \, \, \text{V}$ 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 $F=n\times 10^{- 4}N$ due to the impact of the electrons. The value of $n$ is __________. Mass of the electron $m_{e}=9\times 10^{- 31}kg \, and \, 1.0 \, eV=1.6\times 10^{- 19}J$ .

Answer 1

Power $=Nh\nu$
$N=$ number of photons per second
Since $KE=0,h\nu=\phi$
$200=N\left[6.25 \times 1.6 \times 10^{- 19 \, } \, J o u l e\right]$
$N=\frac{200}{6.25 \times 1.6 \times 10^{- 19}}$
As photon is just above threshold frequency $KE_{m a x}$ is zero and they are accelerated by potential difference of $500 \, V$
$KE_{f}=q∆V$
$\frac{P^{2}}{2 m}=q∆V\Rightarrow P=\sqrt{2 m q ∆ V}$
Since efficiency is 100%, number of electrons emitted is equal to number of photons falling per second as electrons are completely absorbed, force exerted $nmv$
$=\frac{200}{6.25 \times 1.6 \times 10^{-19}} \times \sqrt{2\left(9 \times 10^{-31}\right) \times 1.6 \times 10^{-19} \times 500}$
$= \, \frac{3 \times 200 \times 10^{- 25} \times \sqrt{1600}}{6.25 \times 1.6 \times 10^{- 19}}=\frac{2 \times 40}{6.25 \times 1.6}\times 10^{- 4}\times 3=24\times 10^{- 4}$
$n = 24$