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 at the same rate. The 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 5802 K

• A. the temperature of B is 1934 K
• B. ${\lambda }_B=1.5\mu m$
• 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

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

Power radiated and surface area is same for both A and B .
Therefore, $e_A\sigma T_A^{4}A=e_B\sigma T_B^{4}A$
$\therefore \frac{T_A}{T_B}={\left(\frac{e_B}{e_B}\right)}^{\frac{1}{4}}={\left(\frac{0.81}{0.01}\right)}^{\frac{1}{4}}=3$
$therefore{T}_B=\frac{T_A}{3}=\frac{5802}{3}$
= 1934 K
$T_B=1934K$
According to Wien’ s displacement law,
${\lambda }_{m}T=constant$
$\therefore {\lambda }_A{T}_A={\lambda }_B{T}_B$
or ${\lambda }_A{T}_B={\lambda }_B\left(\frac{T_B}{T_A}\right)=\frac{{\lambda }_B}{3}$
Given, ${\lambda }_B-{\lambda }_A=1\mu m$
$\Rightarrow {\lambda }_B-\frac{{\lambda }_B}{3}=1\mu m$
or $\frac{2}{3}{\lambda }_B=1\mu m$
$\Rightarrow {\lambda }_B=1.5\mu m$

Note: ${\lambda }_{m}T=B$ = Wien’s constant value of this constant for perfectly black body is $2.89\times {10}^3$ m-K. For other bodies this constant will have some different value. In the option (b) has different for different bodies. Option (b) is incorrect.