Question

The power radiated by a black body is \(P\) and it radiates maximum energy at wavelength \(\Lambda_{0}\). If the temperature of the black body is now changed so that it radiates maximum energy at wavelength \(3 / 4 \wedge_{0}\), the power radiated by it becomes \(n \mathrm{P}\). The value of \(n\) is :-
(1) \(3 / 4\)
(2) \(4 / 3\)
(3) \(256 / 81\)
(4) \(81 / 256\)

Answer 1

The correct answer is (3)
$$
\mathrm{P}=\sigma \mathrm{AT}^{4} \Rightarrow \mathrm{P} \propto \mathrm{T}^{4}
$$
According to Wein's law \(\mathrm{T} \propto \frac{1}{\lambda_{\mathrm{m}}}\)
$$
\begin{aligned}
&\Rightarrow \mathrm{P} \propto\left(\frac{1}{\lambda_{\mathrm{m}}}\right)^{4} \Rightarrow \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\lambda_{\mathrm{m}_{1}}}{\lambda_{\mathrm{m}_{2}}}\right)^{4} \\
&\Rightarrow \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\frac{\lambda_{0}}{\frac{3}{4} \lambda_{0}}\right)^{4} \Rightarrow \frac{\mathrm{nP}}{\mathrm{P}}=\frac{256}{81} \Rightarrow \mathrm{n}=\frac{256}{81}
\end{aligned}
$$

Answer 2

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Answer 3

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