Question

In an experiment on photoelectric effect, light of wavelength 400 nm is incident on a cesium plate at the rate of 5.0 W. The potential of the collector plate is made sufficiently positive with respect to the emitter, so that the current reaches its saturation value. Assuming that on average, one out of every 10⁶ photons is able to eject a photoelectron, find the photocurrent in the circuit.

Answer 1

Given:
Wavelength of light, λ = 400 nm
Power, P = 5 W
Energy of photon,
E = $\frac{hc}{\lambda }=\left(\frac{1242}{400}\right)\mathrm{eV}$
Number of photons, n = $\frac{P}{E}$
$n=\frac{5\times 400}{1.6\times {10}^{-19}\times 1242}$

Number of electrons = 1 electron per 10⁶ photons
Number of photoelectrons emitted,
$n\text{'}=\frac{5\times 400}{1.6\times 1242\times {10}^{-19}\times {10}^6}$
Photo electric current,
I = Number of electron $\times$ Charge on electron
$$\begin{gathered} I=\frac{5\times 400}{1.6\times 1242\times {10}^{-19}\times {10}^6}\times 1.6\times {10}^{-19} \\ =1.6\times {10}^{-6}\mathrm{A}=1.6\mathrm{\mu A} \end{gathered}$$