Question

Ultraviolet light of wavelength $300\text{nm}$ and intensity $1{\text{W m}}^{-2}$ falls on the surface of a photosensitive material. If one percent of the incident photons produce photoelectrons then the number of photoelectrons emitted per second from an area of $1{\text{cm}}^2$ of the surface is nearly.
[Take $hc=1240\text{eV (nm)}$]

• A. $1.5\times {10}^{13}$
• B. $1.5\times {10}^{12}$
• C. $4.1\times {10}^{13}$
• D. $2.3\times {10}^{11}$

Answer 1

The correct option is B $1.5\times {10}^{12}$

Given, $\lambda =300\text{nm};I=1{\text{W m}}^{-2}A=1{\text{cm}}^2$

$\because P=Nh\nu =\frac{Nhc}{\lambda }[\because P=IA]$

$\Rightarrow N=\frac{P\lambda }{hc}=\frac{IA\lambda }{hc}$

Where, $N=\text{No. of photons emitted per second}$

$\text{Photon flux}\left(\eta \right)=\frac{\text{No. of photoelectron emitted}}{\text{No. of photons emitted}}$

Here, $\eta =1\%$

$\text{No. of photoelectrons emitted}=1\%\times N$

$=\frac{1}{100}\times \frac{IA\lambda }{hc}$

$=\frac{1}{100}\times \frac{1\times 1\times {10}^{-4}\times 300}{1240\times 1.6\times {10}^{-19}}$

$=\frac{3\times {10}^{12}}{1.984}$

$\approx 1.5\times {10}^{12}$

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