Solved Problem on Black Body Radiation
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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?
Problem data:
- Wavelength at temperature T1: λmax1 = 6500 Å;
- Relationship between spectral radiations at temperatures T1 and T2: RT2 = 2RT1.
First, let's convert the given wavelength in angstrom (Å) to meters (m) used in the International System of Units (SI)
\[
\begin{gather}
\lambda_{max1}=6500\;\cancel{\overset{\circ}{\mathsf{A}}}\times\frac{1\times10^{-10}\;\text{m}}{1\;\cancel{\overset{\circ}{\mathsf{A}}}}=6.5\times 10^{3}\times 10^{-10}\;\text{m}=6.5\times10^{-7}\;\text{m}
\end{gather}
\]
The spectral radiation is given by Stefan-Boltzmann Law
\[
\begin{gather}
\bbox[#99CCFF,10px]
{R_{T}=\sigma T^{4}}
\end{gather}
\]
writing this expression for situations 1 and 2
\[
\begin{gather}
R_{T1}=\sigma T_{1}^{4} \tag{I-a}
\end{gather}
\]
\[
\begin{gather}
R_{T2}=\sigma T_{2}^{4} \tag{I_b}
\end{gather}
\]
substituting the expressions (I-a) and (I-b) into the given relationship between the spectral radiations
\[
\begin{gather}
R_{T1}=2R_{T2}\\[5pt]
\cancel{\sigma} T_{1}^{4}=2\cancel{\sigma} T_{2}^{4}\\[5pt]
T_{1}^{4}=2T_{2}^{4} \tag{II}
\end{gather}
\]
Wien's Law is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\lambda_{max}T=2.989\times 10^{-3}}
\end{gather}
\]
using this expression for situations 1 and 2, we obtain the temperatures
\[
\begin{gather}
\lambda _{max1}T_{1}=2.989\times 10^{-3}\\[5pt]
T_{1}=\frac{2.989\times 10^{-3}}{\lambda_{max1}} \tag{III-a}
\end{gather}
\]
\[
\begin{gather}
\lambda _{max2}T_{2}=2.989\times 10^{-3}\\[5pt]
T_{2}=\frac{2.989\times 10^{-3}}{\lambda_{max2}} \tag{III-b}
\end{gather}
\]
substituting expressions (III-a) and (III-b) into expression (II)
\[
\begin{gather}
\left(\frac{2.989\times 10^{-3}}{\lambda_{max2}}\right)^{4}=2\left(\frac{2.989\times 10^{-3}}{\lambda_{max1}}\right)^{4}\\[5pt]
\frac{\cancel{\left(2.989\times 10^{-3}\right)^{4}}}{\lambda_{max2}^{4}}=2\frac{\cancel{\left(2.989\times 10^{-3}\right)^{4}}}{\lambda_{max1}^{4}}\\[5pt]
\frac{1}{\lambda_{max2}^{4}}=2\frac{1}{\lambda_{max1}^{4}}\\[5pt]
\lambda_{max2}^{4}=\frac{\lambda_{max1}^{4}}{2}\\[5pt]
\lambda_{max2}=\sqrt[{4}]{\frac{\lambda_{max1}^{4}}{2}}\\[5pt]
\lambda_{max2}=\frac{\lambda_{max1}}{\sqrt[{4}]{2}}\\[5pt]
\lambda_{max2}=\frac{6.5\times 10^{-7}}{1.189}\\[5pt]
\lambda_{max2}=5.466\times 10^{-7}\;\text{m}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{\lambda_{max2}=5466\;\overset{\circ}{{\mathsf{A}}}}
\end{gather}
\]
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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 .