MHT CET · Physics · Thermal Properties of Matter
A black body radiates maximum energy at wavelength ' \(\lambda\) ' and its emissive power is \(E\). Now due to change in temperature of that body, it radiates maximum energy at wavelength \(\frac{2 \lambda}{3}\). At that temperature emissive power is
- A \(\frac{51 \mathrm{E}}{8}\)
- B \(\frac{81 \mathrm{E}}{16}\)
- C \(\frac{61 \mathrm{E}}{27}\)
- D \(\frac{71 \mathrm{E}}{19}\)
Answer & Solution
Correct Answer
(B) \(\frac{81 \mathrm{E}}{16}\)
Step-by-step Solution
Detailed explanation
From Wien's Displacement law,
\(\lambda_{\max }=\frac{\mathrm{b}}{\mathrm{T}} \Rightarrow \mathrm{T}=\frac{\mathrm{b}}{\lambda_{\max }}\)
From Stefan-Boltzmann law
\(\mathrm{E}=\sigma \mathrm{T}^4=\sigma\left(\frac{\mathrm{b}}{\lambda_{\max }}\right)^4\)
Let the new emissive power be \(\mathrm{E}^{\prime}\).
\(\begin{aligned}
\therefore \quad E^{\prime} & =\sigma\left(\frac{b}{\frac{2 \lambda_{\max }}{3}}\right)^4 \\
\mathrm{E}^{\prime} & =\frac{81}{16} \mathrm{E}
\end{aligned}\)
\(\lambda_{\max }=\frac{\mathrm{b}}{\mathrm{T}} \Rightarrow \mathrm{T}=\frac{\mathrm{b}}{\lambda_{\max }}\)
From Stefan-Boltzmann law
\(\mathrm{E}=\sigma \mathrm{T}^4=\sigma\left(\frac{\mathrm{b}}{\lambda_{\max }}\right)^4\)
Let the new emissive power be \(\mathrm{E}^{\prime}\).
\(\begin{aligned}
\therefore \quad E^{\prime} & =\sigma\left(\frac{b}{\frac{2 \lambda_{\max }}{3}}\right)^4 \\
\mathrm{E}^{\prime} & =\frac{81}{16} \mathrm{E}
\end{aligned}\)
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