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

The photocurrent in the circuit is $1.6 \mu A$

Explanation:

Step 1:

Given data in the question :

Light Wavelength, λ = 400 nm

Power of photoelectric effect, P = 5 W

Photon Energy, is given by $E=\frac{\mathrm{hc}}{\lambda}=\left(\frac{1242}{400}\right) \mathrm{eV}$

No. of photons  $n=\frac{P}{E}$

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

Electron number = 1 electron for every 106 photons

Amount of Emitted Photo electrons,

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

Step 2:

Photo electric current is given by

I = electron Number  ×   electron charge

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

$I=\frac{2000}{1.6 \times 10^{-13} \times 1242} \times 1.6 \times 10^{-19}$

$I=\frac{2000}{10^{-13} \times 1242} \times 10^{-19}$

$I=\frac{2000}{1242} \times 10^{-6}$

 $=1.6 \times 10^{-6} A=1.6 \mu A$