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 $500$V is applied between the cathode and anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experience 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

Power = nhv n = number of photons per second

Since KE = 0, hv = $\varphi$

$200=n[6.25\times 1.6\times {10}^{-19}Joule]$

$n=\frac{200}{1.6\times {10}^{-19}\times 6.25}$

As photon is just above threshold frequency, $K{E}_{max}$ is zero and they are accelerated by potential difference of $500$V.

$K{E}_f=q△V$

$\frac{P^2}{2m}=q△V\Rightarrow P=\sqrt{2mq△V}$

Since efficiency is $100\%$, number of electrons = number of photons per second.

As photon is completely absorbed, force exerted = nmv

= $\frac{200}{6.25\times 1.6\times {10}^{-19}}\times \sqrt{2(9\times {10}^{-31})\times 1.6\times {10}^{-19}\times 500}$

= $\frac{3\times 200\times {10}^{-25}\times \sqrt{1600}}{6.25\times 1.6\times {10}^{-19}}=\frac{2\times 40}{6.25\times 1.6}\times {10}^{-4}\times 3=24$