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 pm. If the temperature of A is 5802 K.

• A. the temperature of it is 1934 K
• B. $\lambda_B =1.5 \mu m$
• C. the temperature of B is i 1604 K
• D. the temperature of B is 2901 K

Answer 1

Answer: (B)

$\lambda_B =1.5 \mu m$

Answer 2

Pow er radiated and surface area is same for both A and B.
Therefore, $e_A \sigma T_A^4 A=e_B \sigma T_B^4 A$
$\therefore \, \, \, \, \, \frac{T_A}{T_B}=\bigg(\frac{e_B}{e_A}\bigg)^{1/4}=\bigg(\frac{0.81}{0.01}\bigg)^{1/4}=3$
$\therefore \, \, \, \, T_B =\frac{T_A}{3}=\frac{5802}{3}=1934 K$
$\, \, \, \, \, \, \, \, \, \, \, \, \, T_B =1934 K$
According to Wien's displacement law,
\hspace20mm \lambda_m T =constant
$\therefore \, \, \, \, \, \, \, \, \, \lambda_A T_A=\lambda_B T_B \bigg(\frac{T_B}{T_A}\bigg)=\frac{\lambda_B}{3}$
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 x 10$^{-3}$ m-K. For other bodies this constant will have some different value. In the opinion of author option (b) has been framed by assuming b to be constant for all bodies. If we take b different for different bodies. Option (b) is incorrect.