Question

In an experiment on photoelectric effect, Light of wavelength 800 nm (less than threshold wavelength) 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 on the emitter so that the current reaches its saturation value. Assuming that on the average one of every ${10}^6$ photons is able to eject a photoelectron, find the photo current in the circuit.

Answer 1

$\lambda =800nmP=5WEofphoton=\frac{hc}{\lambda }=\frac{1242}{400}ev$

no. of photon = $5\times \frac{800}{1.6}\times {10}^{-19}\times 1242$

No. of electrons = 1 per ${10}^6$ photon

no. of photoelectron emitted =$5\times \frac{800}{1.6}\times 1242\times {10}^{-19}\times {10}^6$

photo electric current = $5\times \frac{800}{1.6}\times 1242\times {10}^{-19}\times {10}^6\times 1.6\times {10}^{-19}$

photo electric current $=3.2\mu A$