Prove Kirchhoff’s law of radiation theoretically.
Solution
Consider an ordinary body O 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 O
Eb = emissive power of perfectly black body B
a = coefficient of absorption of O
e = emissivity of O
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 O = aQ.
Quantity of heat energy emitted per unit area per unit time by body O = E.
Since there is no change in temperature
E = aQ
Q = E/a .…(1)
Quantity of heat absorbed per unit area per unit time by a perfectly black body, B = Q The radiant heat energy emitted per unit time per unit area by a perfectly black body,
B = Eb
Since there is no change in temperature.
Eb = Q .…(2)
From equations (1) and (2),
`E/a=E_b =>E/E_b=a`
But, `E/E_b=e`
a = e
