Question

In a photoelectric experiment a parallel beam of monochromatic light with power of $200W$ is incident on a perfectly absorbing cathode of work function $6.25eV$. 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 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 $1.0eV=1.6\times 10^{-19}J$.

Answer 1

Answer: 24

power $=nhv$
$n=$ number of protons per second
Since $KE=0hc=\varphi$
$200=n[6.25\times 16\times 10^{-19}$ 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 $500V$.
$K{E}_f=q\Delta V$
$\frac{P^2}{2m}=q\Delta V$
$\Rightarrow P=\sqrt{2m\Delta V}$
Since efficiency is $100\%$, number of electrons = number of photons per second As photon is completely absorbed force exerted $=nmv$
$=\frac{200}{6.25\times 1.6\times 10^{-19}}\times \sqrt{2\left(9\times 10^{-31}\right)\times 1.6\times 10^{-19}\times 500}$
$=\frac{3\times 200\times 10^{-25}\times \sqrt{1600}}{6.25\times 1.6\times 10^{-19}}$
$=\frac{2\times 40}{6.25\times 1.6}\times 10^{-4}\times 3=24$