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 wavelength ${\lambda }_B$ corresponding to maximum spectral radiancy in the radiation from B is 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 ${\lambda }_B=1.5\mu m$

To calculate temperature of B

Power of body $A=e_A\sigma T_A^4\times \left(area\right)$

Power of body $B=e_B\sigma T_B^4\times \left(area\right)$

The two powers are equal,

$\therefore e_B\sigma T_B^4\times \left(area\right)=e_A\sigma T_A^4\times \left(area\right)$

or $T_B^4=\left(\frac{e_A}{e_B}\right)T_A^4$ $T_B^4=\left(\frac{0.01}{0.81}\right)\times (5802{)}^4$

or $T_B=1934K$ $\therefore$ Option (a) is correct

d) According to Wien's displacement law,

$\lambda T=constant$ $\therefore {\lambda }_A{T}_A={\lambda }_B{T}_B$

or $\therefore \frac{{\lambda }_A}{{\lambda }_B}=\frac{T_B}{T_A}$ or $\frac{{\lambda }_A}{{\lambda }_B}=\frac{1934}{5802}$

or $\frac{{\lambda }_A}{{\lambda }_B}=\frac{1}{3}$ or ${\lambda }_B=3{\lambda }_A$

Given: ${\lambda }_B-{\lambda }_A=1.0\times {10}^{-6}m$

$3{\lambda }_A-{\lambda }_A={10}^{-6}\left(or\right)2{\lambda }_A={10}^{-6}$

or ${\lambda }_A=0.5\times {10}^{-6}$ and ${\lambda }_B=3\times 0.5\times {10}^{-6}$

or ${\lambda }_B=1.5\times {10}^{-6}m$

Hence option (b) is correct

Options (c) and (d) are incorrect as option (a) is correct

Hence options (a) and (b) are correct.