Question

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 $1\, \mu m$. If the temperature of the body $A$ is $5802\, K$, then value of $\lambda_{B}$ is

• A. $\frac{1}{2} \mu m$
• B. $1\, \mu m$
• C. $2\, \mu m$
• D. $\frac{3}{2} \mu m$

Answer 1

Answer: (D)

We knows from Stefan's law,
$E=e A \sigma T^{4}$
Here, $E_{1}= e _{1} A \sigma T_{1}^{4}$
$E_{2}=e_{2} A \sigma T_{2}^{4}$
so, $E_{1}=E_{2}$
$\therefore e_{1} T_{1}^{4}=e_{2} T_{2}^{4}$
$\Rightarrow T_{2}=\left(\frac{e_{1}}{e_{2}} T_{1}^{4}\right)^{1 / 4}$
$=\left(\frac{1}{81} \times(5802)^{4}\right)^{1 / 4}$
$\Rightarrow T_{B}=1934 K$
From Wein's law, $\lambda_{A T_{A}}=\lambda_{B} T_{B}$
$\Rightarrow \frac{\lambda_{A}}{\lambda_{B}}=\frac{T_{B}}{T_{A}}$
$\Rightarrow \frac{\lambda_{B}-\lambda_{A}}{\lambda_{B}}=\frac{T_{B}-T_{B}}{T_{A}}$
$\Rightarrow \frac{1}{\lambda_{B}}=\frac{5802-1934}{5802}=\frac{3968}{5802}$
$\Rightarrow \lambda_{B}=\frac{3}{2} \mu m$