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Two bodies A and B radiate maximum energy with wavelength difference $4 \mu \mathrm{m}$. The absolute temperature of body $\mathrm{A}$ is 3 times that of $\mathrm{B}$. The wavelength at which body B radiates maximum energy is:
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Verified Answer
The correct answer is:
$6 \mu m$
Concept: According to the Wiens displacement law:
$\lambda_m T=\text { Constant, }$
Given $\left(\lambda_{m B}-\lambda_{m A}\right)=4 \mu \mathrm{m} \quad---(1)$
$T_A=3 T_B \quad---(2)$
Now, using Wens law
$T_A=\lambda_{m B} T_B \quad---(3)$
$\mathrm{Eq}^{\mathrm{n}}$ (2) divided by $\mathrm{Eq}^{\mathrm{n}}$ (3)
$\begin{aligned}
& \frac{1}{\lambda_{m A}}=\frac{3}{\lambda_{m B}} \\
& \Rightarrow \frac{\lambda_{m B}}{\lambda_{m A}}=3 \\
& \Rightarrow \frac{\left(\lambda_{m B}-\lambda_{m A}\right)}{\lambda_{m A}}=2 \\
& \Rightarrow \lambda_{m A}=\frac{1}{2}\left(\lambda_{m B}-\lambda_{m A}\right)
\end{aligned}$
Now, using eq ${ }^{\mathrm{n}} 1$
$\begin{aligned}
& \lambda_{m A}=\frac{1}{2}(4 \mu \mathrm{m})=2 \mu \mathrm{m} \\
& \therefore \lambda_{m B}=4 \mu+\lambda_{m A}=6 \mu \mathrm{m}
\end{aligned}$
$\lambda_m T=\text { Constant, }$
Given $\left(\lambda_{m B}-\lambda_{m A}\right)=4 \mu \mathrm{m} \quad---(1)$
$T_A=3 T_B \quad---(2)$
Now, using Wens law
$T_A=\lambda_{m B} T_B \quad---(3)$
$\mathrm{Eq}^{\mathrm{n}}$ (2) divided by $\mathrm{Eq}^{\mathrm{n}}$ (3)
$\begin{aligned}
& \frac{1}{\lambda_{m A}}=\frac{3}{\lambda_{m B}} \\
& \Rightarrow \frac{\lambda_{m B}}{\lambda_{m A}}=3 \\
& \Rightarrow \frac{\left(\lambda_{m B}-\lambda_{m A}\right)}{\lambda_{m A}}=2 \\
& \Rightarrow \lambda_{m A}=\frac{1}{2}\left(\lambda_{m B}-\lambda_{m A}\right)
\end{aligned}$
Now, using eq ${ }^{\mathrm{n}} 1$
$\begin{aligned}
& \lambda_{m A}=\frac{1}{2}(4 \mu \mathrm{m})=2 \mu \mathrm{m} \\
& \therefore \lambda_{m B}=4 \mu+\lambda_{m A}=6 \mu \mathrm{m}
\end{aligned}$
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