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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Aicharya · Accepted Answer

Concept: According to the Wiens displacement law: $\lambda_m T=	ext { Constant, }$ Given $\left(\lambda_{m B}-\lambda_{m A}ight)=4 \mu \mathrm{m} \quad---(1)$ $T_A=3 T_B \quad---(2)$ Now, using Wiens displacement law $T_A=\lambda_{m B} T_B \quad---(3)$ $\mathrm{Eq}^{\mathrm{n}}$ (2) divided by $\mathrm{Eq}^{\mathrm{n}}$ (3) $\begin{aligned} & \frac{1}{\lambda_{m A}}=\frac{3}{\lambda_{m B}} \ & \Rightarrow \frac{\lambda_{m B}}{\lambda_{m A}}=3 \ & \Rightarrow \frac{\left(\lambda_{m B}-\lambda_{m A}ight)}{\lambda_{m A}}=2 \ & \Rightarrow \lambda_{m A}=\frac{1}{2}\left(\lambda_{m B}-\lambda_{m A}ight) \end{aligned}$ Now, using eq ${ }^{\mathrm{n}} 1$ $\begin{aligned} & \lambda_{m A}=\frac{1}{2}(4 \mu \mathrm{m})=2 \mu \mathrm{m} \ & 	herefore \lambda_{m B}=4 \mu+\lambda_{m A}=6 \mu \mathrm{m} \end{aligned}$

Two bodies A and B radiate maximum energy with wavelength difference $4 \mu \mathrm{m}$. The absolute…

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