Module 5 : MODERN PHYSICS
Lecture 23 : Black Body Radiation
  Example 3
  Estimate the fraction of radiant power of Example 1 which is emitted in the visible region of the spectrum.
Solution
 
According to Planck's radiation formula, the power per unit area is given by
 
$\displaystyle P = \frac{2\pi h}{c^2}\int_{\nu_1}^{\nu_2}\nu^3 \frac{1} {\exp(h\nu/kT )-1}d\nu$
  Substituting $ x=h\nu/kT$, the expression reduces to
 
$\displaystyle P = \frac{2\pi h}{c^2}\left(\frac{kT}{h}\right)^4\int_{x_1}^{x_2} dx x^3\frac{1}{e^x-1}$
  where $ x_1$ and $ x_2$ are respectively the upper and the lower limits of $ x$ corresponding to visible spectrum. Taking $ \nu_1=4.3\times 10^{14}$ Hz and $ \nu_2 = 7.5\times 10^{14}$ Hz, we get $ x_1 = 3.46$ and $ x_2 = 6$. Thus
 
$\displaystyle P$ $\displaystyle =$ $\displaystyle \frac{2\pi h}{c^2}\left(\frac{kT}{h}\right)^4 \int_{3.46}^{6}dx x^3\frac{1}{e^x-1}$
$\displaystyle =$ $\displaystyle 1.126\times 10^7\int_{3.46}^{6}dx x^3\frac{1}{e^x-1}$
 
The integral above has to be done numerically, for instance, by Simpson's method. A crude estimate gives the value of the integral to be approximately 2.41. Thus $ P = 2.7\times 10^7$ watts, which is about 36% of the total emitted radiation.

 Back