Question

Ultraviolet light of wavelength 300 nm and intensity $1.0watt/m^2$ falls on the surface of a photosensitive material. If 1% of the incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of $1.0c{m}^2$ of the surface is nearly

• A. $9.61\times {10}^{14}$ per sec
  
• B. $4.12\times {10}^{13}$ per sec
  
• C. $1.51\times {10}^{12}$ per sec
  
• D. $2.13\times {10}^{11}$ per sec

Answer 1

The correct option is C $1.51\times {10}^{12}$ per sec


Intensity of light
$I=\frac{Watt}{Area}=\frac{nhc}{A\lambda }\Rightarrow$ Number of photon $n=\frac{IA\lambda }{hc}$
$\therefore$ Number of photo electron = $\frac{1}{100}\times \frac{IA\lambda }{hc}$
$=\frac{1}{100}\frac{1\times {10}^{-4}\times 300\times {10}^{-9}}{6.6\times {10}^{-34}\times 3\times {10}^8}=1.5\times {10}^{12}$