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.
Proof : Consider an ordinary body A and perfectly black body B of the same dimension suspended in a uniform temperature enclosure as shown in the figure.
At thermal equilibrium, both the bodies will have the same temperature as that of the enclosure.
Let, E = emissive power of ordinary body A
Eb = emissive power of perfectly black body B
a = coefficient of absorption of A
e = emissivity of A
Q = radiant energy incident per unit time per unit area on each body Quantity of heat absorbed per unit area per unit time by 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 = 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 R, per unit time per unit surface area.
Since there is no change in its temperature, we must have,
Q=Eb ... (2)
From Eqs. (1) and (2), we get,
a = E/Q = E/Eb …….. (3)
From Eq. (3), we get, E/a= Eb
But by definition of coefficient of emission
e =E/Eb …….. (4)
from eq. 3 & 4 we get
e=a
Hence the proof of kirchoff’s law of radiation
