Question

State Kirchhoff's law of radiation and prove it theoretically.

Answer 1

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 $E_b$ 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
$aQ=E...\left(1\right)$
As body $B$ is a perfect blackbody it will absorb the quantity $Q$ per unit time per unit surface area and radiate the quantity $E_b$ per unit surface area.
Since there is no change in its temperature. We must have
$Q=E_b....\left(2\right)$
From Equation $\left(1\right)$ and $\left(2\right)$ we get
$a=\frac{E}{Q}=\frac{E}{E_b}....\left(3\right)$
By definition of coefficient of emission
$e=\frac{E}{E_b}....\left(4\right)$
From Equation $\left(3\right)$ and $\left(4\right)$ we get $a=e$.
Hence, Kirchhoff's law of radiation proved.