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:

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 \({ }^{\mathrm{n}} 1\)
\(\begin{aligned}
& \lambda_{m A}=\frac{1}{2}(4 \mu \mathrm{m})=2 \mu \mathrm{m} \\
& \therefore \lambda_{m B}=4 \mu+\lambda_{m A}=6 \mu \mathrm{m}
\end{aligned}\)