Question

A light source of 320 Watt emit monochromatic light of wavelength $6200\stackrel{\circ }{A}$. If all the emitted photons are made to strike on the metal plate of work function equal to 1.8 eV and quantum yield is 25% then photocurrent observed in ampere will be:

Answer 1

Given,
Wavelength of the photon = $6200\stackrel{\circ }{A}$
Energy of one photon in eV $=\frac{hc}{\lambda \times 1.6\times {10}^{-19}}=\frac{12400}{6200}eV=2eV$
energy of the light source = 320 W or 320 J/s
Number of photons in one second $=\frac{\text{energy of the light}}{\text{energy of the photon}}$
$=\frac{320}{2\times 1.6\times {10}^{-19}}={10}^{21}$

% yield = 25%
$=\frac{25}{100}\times {10}^{21}=\frac{{10}^{21}}{4}$
One photon corresponds to $1.6\times {10}^{-19}C$
So charge per second $=\frac{{10}^{21}}{4}\times 1.6\times {10}^{-19}C$
= 40 C/s or 40 A