Question

Two bodies $A$ and $B$ have thermal emissivities of $0.01$ and $0.81$ respectively. The outer surfaces of the two bodies are the same. The two bodies radiate energy at the same rate. The wavelength ${\lambda }_B$, corresponding to the maximum spectral radiancy in the radiation from $B$, is shifted from the wavelength corresponding to the maximum spectral radiancy in the radiation from $A$ by $1.00\mu m$. If the temperature of $A$ is $5802K$.

• A. the temperature of $B$ is 1934 K
• B. $\lambda =1.5\mu m$
• C. the temperature of $B$ is 11604 K
• D. the temperature of $B$ is 2901 K

Answer 1

Answer: (A), (B)

The correct options are
A the temperature of $B$ is 1934 K
B $\lambda =1.5\mu m$

Let P and A be the power radiated and the surface area of both bodies respectively.

Let $T_A$ and $T_B$ the absolute temperature of A and B respectively.

$P=0.01A\rho T_A^4=0.81A\rho T_B^4$

$\Rightarrow T_A=3T_B$

$\Rightarrow T_B=5802/3=1934K$

as, by wien's displacement law

${\lambda }_A{T}_A={\lambda }_B{T}_B$

$\Rightarrow {\lambda }_B=3{\lambda }_A$

given that, ${\lambda }_B-{\lambda }_A=1\mu m$

${\lambda }_B=1.5\mu m$

So option A and B are correct.