Question

Ultraviolet light of wavelength 300 nm and intensity $1.0W{m}^{-2}$ 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 $1.0c{m}^2$ 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

Answer 1

Answer: (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 the number of electrons emitted
$=\frac{1}{100}\times \frac{10^{-4}}{6.6\times 10^{-19}}$
$=1.5\times 10^{12}{s}^{-1}$