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 $F = n \times 10^{-4} N$ due to the impact of the electrons. The value of $n$ is __________.
Mass of the electron $m_{e} = 9\times10^{-31}$ kg and $1.0 \,eV = 1.6 \times 10^{-19}\,J$.

Answer 1

power $= nhv $
$n =$ number of protons per second
Since $KE =0 \,\,h c=\phi$
$200=n\left[6.25 \times 16 \times 10^{-19}\right.$ joule ]
$ n=\frac{200}{1.6 \times 10^{-19} \times 6.25}$
As photon just above threshold frequency $K E_{\max }$ is zero
and they are accelerated by potential difference of $500\, V$.
$K E_{f}=q \Delta V $
$\frac{P^{2}}{2 m}=q \Delta V $
$\Rightarrow P=\sqrt{2 m \Delta V}$
Since efficiency is $100 \%$, number of electrons = number of photons per second As photon is completely absorbed force exerted $= nmv$
$=\frac{200}{6.25 \times 1.6 \times 10^{-19}} \times \sqrt{2\left(9 \times 10^{-31}\right) \times 1.6 \times 10^{-19} \times 500}$
$=\frac{3 \times 200 \times 10^{-25} \times \sqrt{1600}}{6.25 \times 1.6 \times 10^{-19}}$
$=\frac{2 \times 40}{6.25 \times 1.6} \times 10^{-4} \times 3=24$