Q. Ultraviolet light of wavelength 300 nm and intensity 1.0 $Wm^{- 2}$ 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) $2.13 \times 10^{11}s^{- 1}$
Explanation
(c) Energy incident over $1cm^{2} = 1.0 \times 10^{- 4}$ J;
Energy required to produce photoelectrons
$= 1.0 \times 10^{- 4} \times 10^{- 2} = 10^{- 6}$ J.
Number of photoelectrons ejected = number of photons which can produce photoelectrons = energy required for producing electron/energy of photon.
$= \frac{10^{- 6}}{hc/\lambda} = \frac{10^{- 6} \times 300 \times 10^{- 9}}{6.6 \times 10^{- 34} \times 3 \times 10^{8}}$
$= 1.51 \times 10^{12}\ s^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.