Question

Ultraviolet light of wavelength $300nm$ and intensity $1.0W{m}^{-2}$ falls on the surface of a photosensitive material. If one per cent of the incident photons produces photoelectrons, then the number of photoelectrons emitted from an area of $1.0{cm}^2$ of the surface is nearly:

• A. $9.61\times {10}^{14}{s}^{-1}$
• B. $4.12\times {10}^{13}{s}^{-1}$
• C. $1.51\times {10}^{12}{s}^{-1}$
• D. $2.13\times {10}^{11}{s}^{-1}$

Answer 1

The correct option is C $1.51\times {10}^{12}{s}^{-1}$

Wavelength $=300nm$

Intensity $=1.0W{m}^{-2}$

Intensity $I=\frac{nhc}{A\lambda }$

No. of photons $n=\frac{IA\lambda }{hc}$

Therefore, the number of photoelectrons

$=\frac{1}{100}\times \frac{IA\lambda }{hc}$ ...[since, $1$% photons produce photoelectrons.]

$\begin{array}{c}=\frac{1}{100}\times \frac{1\times {10}^{-4}\times 300\times {10}^{-9}}{6.6\times {10}^{-34}\times 3\times {10}^8}\\ =1.51\times {10}^{12}{s}^{-1}\end{array}$