Question

State and prove Kirchhoff’s law of heat radiation.

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 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

a Q = 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 = E_b  ​....(2)

From Equation (1) and (2) we get

a = E​/Q = E/E_b​  ​....(3)

By definition of coefficient of emission

e = E/E_b​  ​....(4)

From Equation (3) and (4) we get

a = e

Hence, Kirchhoff's law of radiation proved.

Answer 2

Kirchhoff’s law of heat radiation : At a given temperature, the ratio of the emissive power to the coefficient of absorption of a body is equal to the emissive power of a perfect blackbody at the same temperature for all wavelengths.

OR 

For a body emitting and absorbing thermal radiation in thermal equilibrium, the emissivity is equal to its absorptivity.

Theoretical proof: Consider the following thought experiment: An ordinary body A and a perfect black body B are enclosed in an athermanous enclosure as shown in below figure.

According to Prevost’s theory of heat exchanges, there will be a continuous exchange of radiant energy between each body and its surroundings. Hence, the two bodies, after some time, will attain the same temperature as that of the enclosure.

Let a and e be the coefficients of absorption and emission respectively, of body A. Let R and R_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 aQ per unit time per unit surface area and radiate the quantity R per unit time per unit surface area. Since there is no change in its temperature we must have,

aQ = R … (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 R per unit time per unit surface area.

Since there is no change in its temperature, we must have,

Q = R_b ….. (2) 

From Eqs. (1) and (2), we get,

By definition of coefficient of emission,

\(\cfrac{R}{R_b}\) ......(4)

From Eqs. (3) and (4), we get, a = e. Hence, the proof of Kirchhoff ‘s law of radiation.