Black Body Radiation

Show that the constant between spectral radiance R(ν) and energy density ρ(ν) is c/4.

Use the relationship \( R(\nu)\,d\nu =\frac{c}{4}\rho (\nu )\,d\nu \) between spectral radiance and energy density, and Planck's Radiation Law to derive Stefan's law. That is, show that
\[ R_{T}=\int_{{0}}^{{\infty}}{\frac{2\pi h}{c^{2}}\frac{\nu^{3}\;d\nu }{\operatorname{e}^{h\nu /{kT}}-1}}=\sigma T^{4} \]
where \( \sigma =\dfrac{2\pi ^{5}k^{4}}{15c^{2}h^{3}} \)

hint \( {\Large\int}_{{0}}^{{\infty}}{\dfrac{q^{3}\;dq}{\operatorname{e}^{q}-1}}=\dfrac{\pi^{4}}{15} \)

a) Determine the rest mass lost per second by the sun in the form of radiation. Data: sun surface temperature: 5700 K; sun diameter: 1.4 × 109 m; Stefan-Boltzmann constant: \( \sigma =5.67\times 10^{-8}\;\frac{\text{W}}{\text{m}^{2}.\text{T}^{4}} \); speed of light in vacuum: 3.0 × 108 m/s;
b) What is the fraction of the rest mass lost each year by the sun from the electromagnetic radiation. Solar mass: 2.0 × 1030 kg.

A cavity radiator at 6000 K has a 0.10 mm diameter hole drilled into its wall. Find the power radiated through the hole in the wavelength range from 5500 Å to 5510 Å.
Solution using frequency

Solution using wavelength

At a given temperature λmax = 6500 Å for a blackbody cavity. What will λmax be if the temperature of the cavity walls is increased so that the rate of emission of spectral radiation is doubled?