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 1.00μm. If the temperature of A is 5802 K.
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

Hint: It is given that surface area is the same for both the bodies. The two bodies emit the same power. We have to find the wavelength and corresponding temperature. We use the equation for power radiated by a body and Wien’s law to solve this problem.

Formula used:
We use equation for power radiated:
Power radiated, $P=\sigma eA{{T}^{4}}$
Where A is the surface area of the body, T is the temperature of the body, $e$ is the emissivity of the body and $\sigma $ is the Stefan-Boltzmann constant.
We also use Wien’s law,
${{\lambda }_{B}}=\dfrac{b}{{{T}_{B}}}$
Where b=0.3cm kelvin and ${{\lambda }_{B}}$ is the wavelength and T is the temperature.

Complete step by step solution:
There are two bodies A and B with the same surface area and power radiated by both the bodies is the same.
Emissivity of body A, ${{e}_{A}}=0.01$
Emissivity of body B, ${{e}_{B}}=0.81$
We have to find the wavelength and corresponding temperature of body B. Shift in the wavelength is 1.00 μm.
Temperature of body A, ${{T}_{A}}=5802\,K$

First, we consider the statement that power radiated and surface area of both the bodies is same. That is:
$\sigma {{e}_{A}}A{{T}_{A}}^{4}=\sigma {{e}_{B}}A{{T}_{B}}^{4}$
From which on further solving we get:
\[\dfrac{{{T}_{A}}}{{{T}_{B}}}={{\left( \dfrac{{{e}_{A}}}{{{e}_{B}}} \right)}^{\dfrac{1}{4}}} \\ \]
On equating values, we get temperature of second body as:
${{T}_{B}}=1934\,K$
Now applying Wien’s displacement law,
${{\lambda }_{B}}=\dfrac{b}{{{T}_{B}}}=\dfrac{0.3cmK}{1934K}=0.000155cm=1.5\,\mu m$

Therefore, the answer is options A and B.

Note: This question has two answers. If we consider b different for different bodies then we get another answer. This question can also be solved without taking b value but with the equation ${{\lambda }_{B}}{{T}_{B}}={{\lambda }_{A}}{{T}_{A}}$ which is another form of Wien’s law and shift in wavelength is given.