Question

Ultraviolet light of wavelength $300nm$ and intensity $1.0W{m}^{-2}$ falls on the surface of a photosensitive material. If one percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted 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 $\lambda =300nm=300\times {10}^{-9}m,$

Intensity, $I=1W{m}^{-2}I=\frac{nhc}{\lambda }\to n=\frac{I\lambda }{hc}=\frac{1\times 300\times {10}^{-9}}{6.6\times {10}^{-34}\times 3\times {10}^8}=1.5\times {10}^{18}1$

Now $n_1$ photons are incident on $4\pi r^2=1㎡$area

$\therefore n^{\prime \prime }photonsareincidenton1{cm}^2={10}^{-4}{m}^{2}areaare\frac{1.5\times {10}^{16}}{n^{\prime \prime }}=\frac{1}{{10}^{-4}}\to n^{\prime \prime }=1.5\times {10}^{12}/s$