Module 5 : MODERN PHYSICS
Lecture 24 : Photoelectric Effect
  Example 9
A monochromatic source of light with a wavelength 200 nm and power output of 2 watts is held at a distance of 0.1m from the surface of an aluminium foil. Aluminium has a work function of 4.2 eV and an atomic radius of 0.15 nm. Take the photo-emission efficiency to be 2.5%. Calculate
the kinetic energy of the fastest and the slowest photoelectron emitted,
the average number of photons falling on an atom of Al per second,
the number of photoelectrons emitted per unit area per second.
  Solution
200 nm corresponds to a photon energy of $ 9.95\times 10^{-19}$ J, which is equal to 6.2 eV. Thus the kinetic energy
  of the fastest electron is $ h\nu-\varphi = 6.2-4.2 = 2$ eV. The kinetic energy of the slowest electron is zero.
The light falling on an unit area at a distance $ D$ from the source has an intensity of
  $ I/4\pi D^2 = 2/4\pi(0.1)^2 = 15.92 $ W/m $ ^2$. The amount of radiation captured by an atom of radius $ r$ is
 
  The number of photons is obtained by dividing this by the energy of a single photon. Thus the number of photons captured by an atom on the foil is $ 4.5\times 10^{-18}/9.95\times 10^{-19} = 4.5$ per second.
The photo-emission efficiency is the ratio of the number of photoelectrons emitted from a surface to the number
  of photons falling on the surface in a given time. The number of electrons falling on unit area of the foil is $ 15.92/9.95\times 10^{-19} = 1.6\times 10^{19}$ per second. With 2.5% efficiency, the number of electrons emitted is $ 4\times 10^{17}$ /m $ ^2$-s.

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