State Kirchhoff's law of radiation and prove it theoretically.
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Solution
Kirchhoff's law of radiation : At a given temperature the coefficient of absorption of a body is equal to its coefficient of emission. Theoretical proof : Consider the following thought experiment. An ordinary body A and a perfectly black body B are enclosed in an athermanous enclosure as shown in figure. According to the theory of heat exchanges there will be a continuous exchange of heat energy between each body and its surroundings. Hence, the two bodies, after some time, will attain the same temperature as that of enclosure. Let a and e be the coefficients of absorption and emission respectively of body A and body B. Let E and Eb be the emissive powers of bodies A and B respectively. Suppose that Q is the quantity of radiant energy incident on each body per unit time per unit surface area of the body. Body A will absorb the quantity a Q per unit time per unit surface area and radiate the quantity E per unit time per unit surface area. Since there is no change in its temperature, we must have aQ=E...(1) As body B is a perfect blackbody it will absorb the quantity Q per unit time per unit surface area and radiate the quantity Eb per unit surface area. Since there is no change in its temperature. We must have Q=Eb....(2) From Equation (1) and (2) we get a=EQ=EEb....(3) By definition of coefficient of emission e=EEb....(4) From Equation (3) and (4) we get a=e. Hence, Kirchhoff's law of radiation proved.