Q. Ultraviolet light of wavelength 300 nm and intensity $1.0\ {Wm}^{- 2}$ falls on the surface on photoelectric metal. If one percent of incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of $1.0\ {cm}^{2}$ of the surface is nearly
A) $2.13 \times 10^{11}\ s^{- 1}$
B) $1.5 \times 10^{12}\ s^{- 1}$
C) $3.02 \times 10^{12}\ s^{- 1}$
D) None of these
Explanation
(b) Energy of each photon =$\ \ \ \frac{hc}{\lambda}$
$= \ \ \frac{6.6 \times 10^{- 34} \times 3 \times 10^{8}}{300 \times 10^{- 9}}$
$= \ 6.6 \times 10^{- 19}\ J$
Power of source is given by
$p =$ intensity $\times$ area
$= \ 1.0 \times 1.0 \times 10^{- 4}\ W$
$= \ 10^{- 4}W$
No. of photons per second
$= \ \frac{p}{e} = \frac{10^{- 4}}{6.6 \times 10^{- 19}}$
Now number of electron emitted
$= \ \frac{1}{100} \times \frac{10^{- 4}\frac{}{}}{6.6 \times 10^{- 19}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1.5 \times 10^{12}\ s^{- 1}\ \ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.