Module 5 : MODERN PHYSICS
Lecture 24 : Photoelectric Effect
  Example 10
  Radiation from a black body at 6000 K strikes the surface of a metal with work function 2 eV. What fraction of the black body's total radiant intensity is effective in producing photoelectrons ?
Solution
  The work function 2 eV corresponds to a threshold wavelength $ \lambda_{th} = 621$ nm. Thus irradiance due to very small wavelengths up to 621 nm will cause photoelectrons to be emitted. Using the expression for the radiant intensity, the fraction of the total intensity is
 
$\displaystyle f = \frac{\int_{x_0}^\infty \frac{x^3 dx}{e^x-1}} {\int_0^\infty \frac{x^3 dx}{e^x-1}}$
  where $ x_0 = hc/\lambda_{th}kT= 3.87$. The integral in the numerator has to be done numerically while the value of the denominator is known to be $ \pi^4/15 = 6.494$. The value of the numerator is found to be 2.775, which gives the fraction to be 0.43.
 

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