justify the statement-\"A good absorber is a good emitter\".

Dear Pramod,

Perhaps the most fundamental conceptual way to approach this question is to observe that a hot object placed in a room must ultimately come to thermal equilibrium with the room. The hot object will initially emit more energy into the room than it absorbs from the room, but that will cause the temperature of the room to rise and the temperature of the object to drop. But when they reach the same temperature, we can conclude that the amount of energy absorbed on average is exactly the same as the energy emitted.

Consider a hollow enclosure with its walls maintained at temperature T. The enclosure is filled with thermal radiation. Let I be the energy per unit time falling per unit area of any (imaginary) surface in the enclosure. I is called irradiance in the enclosure. Imagine now that a black body at the same temperature T is introduced in the enclosure. The body will be in thermal equilibrium, i.e., the radiant energy per unit time per unit area emitted by the black body (E_B) equals the radiant energy per unit time per unit area absorbed by it. That is,

since a_B = 1 for the black body.

Next, introduce a non-black body (a < 1) at the temperature T inside the enclosure. Thermal equilibrium now demands that

E = a I

Where E is the radiant energy per unit time per unit area emitted by the (non-black) body. It follows from Eq.(1) that

But in terms of emissivity e already defined

which shows

a = e

That is, the absorptivity of a body equals its emissivity. This is known as Kirchhoffs Law. A good absorber is thus a good emitter.

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