Module 5 : MODERN PHYSICS
Lecture 23 : Black Body Radiation
$\displaystyle \frac{dI(\lambda)}{d\lambda}$ $\displaystyle =$ $\displaystyle (2\pi hc^2)\frac{\partial}{\partial\lambda} \left[\frac{1}{\lambda^5}\frac{1}{\exp(hc/\lambda kT) -1}\right]$
$\displaystyle =$ $\displaystyle (2\pi hc^2)\left[\frac{-5}{\lambda^6}\frac{1}{\exp(hc/\lambda kT)... ...mbda kT)\left(\frac{-hc}{kT\lambda^2}\right)}{(\exp(hc/\lambda kT)-1)^2}\right]$
$\displaystyle =$ 0
  which gives
 
$\displaystyle \frac{hc}{kT\lambda}\frac{1}{1-\exp(-hc/\lambda kT)}=5$
  This equation is to be solved numerically. Substituting $ x= hc/kT\lambda$, the equation becomes
 
$\displaystyle 5(1-e^{-x}) = x$
 
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