Question

Two bodies $A$ and $B$ have thermal emissivities of $0.01$ and $0.81$ respectively the outer surface area of the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelengths ${\lambda }_A$ and ${\lambda }_B$ corresponding to maximum spectral radiancy in the radiation from $A$ and $B$ respectively differ by $1.00\mu m$. If the temperature $A$ is $5802K$, Find
(a) the temperature of $B$
(b) ${\lambda }_B$.

Answer 1

Given: $e_A=0.01,e_B=0.81$ and $T_A=5802K$
From Wien's displacement law
${\lambda }_{m}T=constant\therefore {\lambda }_A{T}_A={\lambda }_B{T}_B$
Power radiated, $P=e\sigma T^{4}A$ as $P_1=P_2$ and $A_1=A_2$
We have $e_A{T}_A^4=e_B{T}_B^4$
$\therefore T_B={\left(\frac{e_A}{e_B}\right)}^{1/4}{T}_A={\left(\frac{0.01}{0.81}\right)}^{1/4}\times 5802=1934K$
As $T_B<T_A,{\lambda }_B>{\lambda }_A$
$\therefore {\lambda }_B-{\lambda }_A=1\mu m$ (given)
${\lambda }_B-{\lambda }_A=1\times {10}^{-6}m.....\left(i\right)$
and $\frac{{\lambda }_B}{{\lambda }_A}=\frac{T_A}{T_B}=\frac{5802}{1934}=3$
$\therefore {\lambda }_B=3{\lambda }_A....\left(ii\right)$
From eq. (i) and (ii), ${\lambda }_B=1.5\times {10}^{-5}m$.