Question

A beam of light of wavelength $400\, nm$ and power $1.55\, mW$ is directed at the cathode of a photoelectric cell. If only $10 \%$ of the incident photons effectively produce photoelectron, then find current due to these electrons. (given, $h c=1240 \,eV - nm , e=1.6 \times 10^{-19} C$ )

• A. $5\,\mu A$
• B. $40 \,\mu A$
• C. $50\,\mu A$
• D. $11.4\,\mu A$

Answer 1

Answer: (C)

The energy of incident photon of wavelength $400 \,nm$ is
$E=\frac{h c}{\lambda}=\frac{1240}{400}=3.1 \,eV$
$\therefore $ Number of such photons in a beam of light of power $1.55 \,mW$
$=\frac{1.55 \times 10^{-3}}{3.1 \times 1.6 \times 10^{-19}} $
$=3.125 \times 10^{15} $ per second
This implies, number of photons that were used to produce photoelectron per second
$=10 \%$ of $3.125 \times 10^{15}$
$=3.125 \times 14$ per second
Hence, the current due to such photoelectron per second
$=3.125 \times 10^{14} \times 1.6 \times 10^{-19} $
$=5 \times 10^{-5} A =50\, \mu A$