Question

In an experiment on photoelectric effect, light of wavelength $400\text{nm}$ is incident on a caesium plate at the rate of $5.0\text{W}$. 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 out of every ${10}^6$ photons is able to eject a photoelectron, find the photocurrent in the circuit.
[Use $hc=1240\text{eV (nm)}$]

• A. $3.2\mu \text{A}$
• B. $1.6\mu \text{A}$
• C. $4.8\mu \text{A}$
• D. $6\mu \text{A}$

Answer 1

The correct option is B $1.6\mu \text{A}$

$\text{Given,}\lambda =400\text{nm};P=5\text{W}$

As we know, Power of a source is,

$P=nh\nu \Rightarrow n=\frac{P}{h\nu }....\left(1\right)$

$\text{Where,}n=\text{No. of photons emitted per second}h\nu =\text{Energy of a photon}$

Now, Energy of a photon is,

$E=h\nu =\frac{hc}{\lambda }=\frac{1240}{400}\text{eV}$

Using, (1) we get,

$n=\frac{P}{h\nu }=\frac{5\times 400}{1.6\times {10}^{-19}\times 1240}$

Given,

${10}^6\text{photons ejects}\to 1\text{photoelectron}$

Now, no.of photoelectrons emitted per second is,

$\frac{N}{t}=\frac{n}{{10}^6}=\frac{5\times 400}{1.6\times {10}^{-19}\times 1240\times {10}^6}$

Photo electric current is,

$i=\frac{Ne}{t}=\frac{5\times 400\times 1.6\times {10}^{-19}}{1.6\times {10}^{-19}\times 1240\times {10}^6}$

$=1.6\times {10}^{-6}\text{A}=1.6\mu \text{A}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.