In an experiment on photoelectric effect, light of wavelength 800 nm (less than threshold wavelength

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 to 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, the photocurrent (in $\mu \mathrm{A}$ ) in the circuit is $\_\_\_\_$ .

Professor PC Thomas Classes · Answer

(3.2): Number of photons falling in one second $=\frac{P}{E_\lambda}=\frac{P \lambda}{h c}$ where $P$ is power of light and $E_\lambda$ is energy of photon. Number of photoelectrons emitted per second $=\frac{P \lambda}{h c} \cdot \frac{1}{10^6}$ $ \begin{aligned} & 	herefore \quad 	ext { Photocurrent }=\frac{P \lambda}{h c 	imes 10^6} \cdot e \ & =\frac{5 	imes 800 	imes 10^{-9} 	imes\left(1.6 	imes 10^{-19}ight)}{6.63 	imes 10^{-34} 	imes 3 	imes 10^8 	imes 10^6} \mathrm{~A}=3.2 \mu \mathrm{~A} \end{aligned} $