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

Answer 1

Power $=nhv$; $n_0=$ number of photons per second since
$KE=0,hv=\varphi$

$200=n_0[6.25\times 1.6\times {10}^{-19}\text{Joule}]$

$n_0=\frac{200}{1.6\times {10}^{-19}\times 6.25}$

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

$K{E}_f=q\Delta V$

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

since efficiency is $100\%,$ number of electrons = number of photons per second.
As photon is completely absorbed, force exerted = $n_{0}mv=n_{0}P$

$\therefore$ The force experienced by the anode is
$F=NP=2\times {10}^{20}\times 1.2\times {10}^{-23}$
$=24\times {10}^{-4}N$

$\therefore$ n = 24