Question

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

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

Answer 1

Answer: (A), (B)

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

Energy emitted per second by body $A={ϵ}_A\sigma T_A^{4}A$
where $A$ is the surface area.
Energy emitted per second by body $B={ϵ}_B\sigma T_B^{4}A$
Given that power radiated is equal
${ϵ}_A\sigma T_A^{4}A={ϵ}_B\sigma T_B^{4}A,{ϵ}_A{T}_A^4={ϵ}_B{T}_B^4$
$\Rightarrow T_B={\left(\frac{{ϵ}_A}{{ϵ}_B}\right)}^{1/4}{T}_A=1934K$
According to Wien's displacement law $\left({\lambda }_m\right)\propto \frac{1}{T}$
Since temperature of $A$ is more, therefore $({\lambda }_m{)}_A$ is less
$\therefore ({\lambda }_m{)}_B-({\lambda }_m{)}_A=1\times {10}^{-6}m$ (given) ....(i)
Also according to Wien's displacement law
$({\lambda }_m{)}_A{T}_A=({\lambda }_m{)}_B{T}_B$
$\Rightarrow \frac{({\lambda }_m{)}_A}{({\lambda }_m{)}_B}=\frac{T_B}{T_A}=\frac{1934}{5802}=\frac{1}{3}....\left(ii\right)$
On solving Eqs. (i) and (ii),
we get ${\lambda }_B=1.5\times {10}^{-6}m$.