Question

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

Answer 1

Answer: (B)

Energy of each photon = $ \, \, \, \frac{h c}{\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} \, \, $