Question

In an experiment on photo electric 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 the average one out of every ${10}^6$ photons is able to eject a photo electron, find the photo current in the circuit.

Answer 1

Given $\lambda =400nm,$

$P=5W$

E of 1 photon $=\frac{hc}{\lambda }=\left(\frac{1242}{400}\right)\text{eV}$

Number of electrons

$\frac{5}{\text{Energy of 1photon}}$

$=\frac{5\times 400}{1.6\times {10}^{-19}\times 1242}$

Number of electrons $=1\text{per}{10}^6\text{photon}$

No. of photo electrons emitted $=\frac{5\times 400}{1.6\times 1242\times {10}^{-19}\times {10}^6}$

Photo electric current

$=\frac{5\times 400}{1.6\times 1242\times {10}^{-19}\times {10}^6}\times 1.6\times {10}^{-19}$

$=1.6\times {10}^{-6}A=1.6\mu A$