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 $6.25eV$ . 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 $500\text{V}$ 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 $m_e=9\times 10^{-31}kgand1.0eV=1.6\times 10^{-19}J$ .

Answer 1

Answer: 24

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