Black Body Radiation
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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
hint \( {\Large\int}_{{0}}^{{\infty}}{\dfrac{q^{3}\;dq}{\operatorname{e}^{q}-1}}=\dfrac{\pi^{4}}{15} \)
\[
R_{T}=\int_{{0}}^{{\infty}}{\frac{2\pi h}{c^{2}}\frac{\nu^{3}\;d\nu }{\operatorname{e}^{h\nu /{kT}}-1}}=\sigma T^{4}
\]
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.
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.
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?
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .