Module 5 : MODERN PHYSICS
Lecture 23 : Black Body Radiation
  Example 1
  Find the temperature for which the radiant energy density at a wavelength of 200 nm is four times that of the density at 400 nm.
  Solution
 
$\displaystyle \frac{u(\lambda_1}{u(\lambda_2)} = \left(\frac{\lambda_2}{\lambda_1}\right)^5 \frac{\exp(hc/\lambda_2kT)-1}{\exp(hc/\lambda_1kT)-1} = 4$
  Substituting values of $ h,c,k, \lambda_1$ and $ \lambda_2$, we get
 
$\displaystyle 32\frac{\exp(3.6\times 10^4/T)-1}{\exp(7.2\times 10^4/T)-1} = 4$
  which gives, on simplification $ \exp(3.6\times 10^4/T) =7$. On solving, the temperature works out approximately to be 18,500 K.

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