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Two bodies $A$ and $B$ of equal surface area have thermal emissivities of 0.01 and 0.81 respectively. The two bodies are radiating energy at the same rate. Maximum energy is radiated from the two bodies $A$ and $B$ at wavelengths $\lambda_A$ and $\lambda_B$ respectively. Difference in these two wavelengths is $1 \mu \mathrm{m}$. If the temperature of the body $A$ is $5802 \mathrm{~K}$, then value of $\lambda_B$ is
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Verified Answer
The correct answer is:
$\frac{3}{2} \mu \mathrm{m}$
We knows from Stefan's law,
Here,
$$
E=e A \sigma T^4
$$
$$
E_1=e_1 A \sigma T_1^4
$$
$$
E_2=e_2 A \sigma T_2^4
$$
so,
$$
E_1=E_2
$$
$$
\begin{array}{lll}
\therefore & e_1 T_1^4=e_2 T_2^4 \\
\Rightarrow & T_2=\left(\frac{e_1}{e_2} T_1^4\right)^{1 / 4}=\left(\frac{1}{81} \times(5802)^4\right)^{1 / 4} \\
\Rightarrow & T_B=1934 \mathrm{~K}
\end{array}
$$
From Wein's law, $\lambda_{A T_A}=\lambda_B T_B$
$\begin{array}{llrl}\Rightarrow & \frac{\lambda_A}{\lambda_B} & =\frac{T_B}{T_A} \\ \Rightarrow & \frac{\lambda_B-\lambda_A}{\lambda_B} & =\frac{T_B-T_B}{T_A} \\ \Rightarrow & \quad \frac{1}{\lambda_B} & =\frac{5802-1934}{5802}=\frac{3968}{5802} \\ \Rightarrow & \lambda_B & =\frac{3}{2} \mu \mathrm{m}\end{array}$
Here,
$$
E=e A \sigma T^4
$$
$$
E_1=e_1 A \sigma T_1^4
$$
$$
E_2=e_2 A \sigma T_2^4
$$
so,
$$
E_1=E_2
$$
$$
\begin{array}{lll}
\therefore & e_1 T_1^4=e_2 T_2^4 \\
\Rightarrow & T_2=\left(\frac{e_1}{e_2} T_1^4\right)^{1 / 4}=\left(\frac{1}{81} \times(5802)^4\right)^{1 / 4} \\
\Rightarrow & T_B=1934 \mathrm{~K}
\end{array}
$$
From Wein's law, $\lambda_{A T_A}=\lambda_B T_B$
$\begin{array}{llrl}\Rightarrow & \frac{\lambda_A}{\lambda_B} & =\frac{T_B}{T_A} \\ \Rightarrow & \frac{\lambda_B-\lambda_A}{\lambda_B} & =\frac{T_B-T_B}{T_A} \\ \Rightarrow & \quad \frac{1}{\lambda_B} & =\frac{5802-1934}{5802}=\frac{3968}{5802} \\ \Rightarrow & \lambda_B & =\frac{3}{2} \mu \mathrm{m}\end{array}$
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