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 $5802K$, 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,
$E=eA\sigma T^4$
Here, $E_1=e_{1}A\sigma T_1^4$
$E_2=e_{2}A\sigma T_2^4$
so, $E_1=E_2$
$\therefore e_1{T}_1^4=e_2{T}_2^4$
$\Rightarrow T_2={\left(\frac{e_1}{e_2}{T}_1^4\right)}^{1/4}$
$={\left(\frac{1}{81}\times (5802{)}^4\right)}^{1/4}$
$\Rightarrow T_B=1934K$
From Wein's law, ${\lambda }_{A{T}_A}={\lambda }_B{T}_B$
$\Rightarrow \frac{{\lambda }_A}{{\lambda }_B}=\frac{T_B}{T_A}$
$\Rightarrow \frac{{\lambda }_B-{\lambda }_A}{{\lambda }_B}=\frac{T_B-T_B}{T_A}$
$\Rightarrow \frac{1}{{\lambda }_B}=\frac{5802-1934}{5802}=\frac{3968}{5802}$
$\Rightarrow {\lambda }_B=\frac{3}{2}\mu m$