Question

Ultraviolet light wavelength $300 \,nm$ and intensity $1.0 \,Wm ^{-2}$ falls on the surface of a photoelectric material. If one percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted per second from an area of $1.0 \,cm ^{2}$ of the surface is nearly

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

Answer 1

The Correct Option is C

Energy of each photon,
$E=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{300 \times 10^{-9}}$
$E=6.6 \times 10^{-19} J$
Power of source
$P=I A=1.0 \times 1.0 \times 10^{-4}=10^{-4} W$
So, number of photons per sec as
$\frac{N}{t}=\frac{P}{E}=\frac{10^{-4}}{6.6 \times 10^{-19}} $
Number of electrons emitted is
$N' =\frac{1}{100} \times \frac{10^{-4}}{6.6 \times 10^{-19}}$
$\Rightarrow N '=1.51 \times 10^{12} / s$