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

• A. $1.52\times {10}^{13}$
• B. $1.51\times {10}^{12}$
• C. $4.12\times {10}^{13}$
• D. $2.13\times {10}^{11}$

Answer 1

The correct option is B $1.51\times {10}^{12}$

Given that $\lambda =300\text{nm},I=1.0{\text{W/m}}^2$

Intensity is given by formula

$I=n\frac{hc}{\lambda }\text{Here n is number of incident photons}\Rightarrow n=\frac{I\lambda }{hc}=1.51\times {10}^{18}{\text{m}}^{-2}{\text{s}}^{-1}$
Number of photons emitted is
$n^1=1\%\text{of}n=1.51\times {10}^{16}{\text{m}}^{-2}{\text{s}}^{-1}$
Number of photons emitted per second from area of $1{\text{cm}}^2$ is $=1.51\times {10}^{16}\times {10}^{-4}\times 1=1.51\times {10}^{12}$