A monochromatic point source $S$radiating wavelength $6000\stackrel{\circ }{A}$, with power 2 watt, an aperture $A$ of diameter $0.1m$ and a large screen $SC$ are placed as hown in figure. A photoemissive detector $D$of surface area $0.5c{m}^2$ is placed at the centre of the screen. The efficiency of the detector for the photoelectron generation for incident photons is $0.9$.

 Calculate the  photocurrent in the detector.

 (a) Energy of one photon,

$E=\frac{hc}{\lambda }=\frac{\left(6.6\times {10}^{-34}\right)\left(3\times {10}^8\right)}{6000\times {10}^{-10}}$

$E=3.3\times {10}^{-19}J$

Power of the source is $2W=2J{s}^{-1}$. Therefore, number of photons emitting per second is

$n_1=\frac{2}{3.3\times {10}^{-19}}=6.06\times {10}^{18}{s}^{-1}$

At distance $0.6m$ , number of photons incident per unit area per unit time is

$n_2=\frac{n_1}{4\pi (0.6{)}^2}=1.34\times {10}^{18}{m}^{-2}{s}^{-1}$

Area of aperture is,

$S_1=\frac{\pi }{4}{d}^2=\frac{\pi }{4}(0.1{)}^2=7.85\times {10}^{-3}{m}^2$

So, total number of photons incident per unit time on the aperture,

$n_3=n_2{S}_1=\left(1.34\times {10}^{18}\right)\left(7.85\times {10}^{-3}\right)s^{-1}$

$\Rightarrow n_3=1.052\times {10}^{16}{s}^{-1}$

This aperture will become new source of light. Now these photons are further distributed in all directions. Hence, at the location of the detector, photons incident per unit area per unit time is

$n_4=\frac{n_3}{4\pi (6-0.6{)}^2}=\frac{1.052\times {10}^{16}}{4\pi (5.4{)}^2}$

$\Rightarrow n_4=2.87\times {10}^{13}{m}^{-2}{s}^{-1}$

This is the photon flux at the centre of the screen. Area of detector is $0.5c{m}^2$ or $0.5\times {10}^{-4}{m}^2$. Therefore, total number of photons incident on the detector per unit time is

$n_5=\left(0.5\times {10}^{-4}\right)\left(2.87\times {10}^{13}d\right)=1.435\times {10}^9{s}^{-1}$

The efficiency of photoelectron generation is $0.9$. Hence, total photoelectrons generated per unit time is

$n_6=0.9n_5=1.2915\times {10}^9{s}^{-1}$

Hence, photocurrent in the detector is

$i=\left(e\right)n_6=\left(1.6\times {10}^{-19}\right)\left(1.2915\times {10}^9\right)$

$\Rightarrow i=2.07\times {10}^{-10}A$