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

On solving Eqs. (i) and (ii)
we get ${\lambda }_B=1.5\times {10}^{-6}m$