A light beam of wavelength 600 nm and intensity 3.3 x 10−3 J m−2 s−1 falls normally on a photocathode of surface area 10−4 m2 and work function 3.2 x 10−19 J. If there is no loss of light by reflection, the number of photo-electrons emitted per second will be:
(Take h = 6.6 x 10−34 J s)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1012
In the photon picture, each photon that ejects an electron gives one photoelectron, so the number of electrons per second equals the number of photons per second (taking every absorbed photon to eject an electron).
Power falling on the cathode:
\( \displaystyle P = IA = 3.3\times10^{-3}\,\mathrm{J\,m^{-2}\,s^{-1}} \times 10^{-4}\,\mathrm{m^{2}} = 3.3\times10^{-7}\,\mathrm{W} \)
Energy of one photon (\( c = 3\times10^{8}\,\mathrm{m\,s^{-1}} \)):
\( \displaystyle E = \frac{hc}{\lambda} = \frac{6.6\times10^{-34} \times 3\times10^{8}}{600\times10^{-9}}\,\mathrm{J} = 3.3\times10^{-19}\,\mathrm{J} \)
Since \( E = 3.3\times10^{-19}\,\mathrm{J} \) is greater than the work function \( 3.2\times10^{-19}\,\mathrm{J} \), emission does take place.
\( \displaystyle n = \frac{P}{E} = \frac{3.3\times10^{-7}}{3.3\times10^{-19}}\,\mathrm{s^{-1}} = 10^{12}\,\mathrm{s^{-1}} \)
The work function only decides whether emission occurs; it does not change the number of electrons.