Q.State and prove Kirchhoff's law of heat radiation.

Good absorbers of radiation are correspondingly good emitters — this proportionality is exactly the black-body emissive power.

Statement: At a given temperature, for radiation of a particular wavelength, the ratio of the emissive power $e_\lambda$ to the absorptive power $a_\lambda$ is the same for all bodies and is equal to the emissive power $E_\lambda$ of a perfectly black body at that temperature:

$\dfrac{e_\lambda}{a_\lambda} = E_\lambda \;(\text{same for all bodies at that } T)$

Proof: Consider a body placed inside an enclosure whose walls are maintained at temperature $T$, in thermal equilibrium with the enclosure. Let the body have emissive power $e_\lambda$ and absorptive power $a_\lambda$, and let $E_\lambda$ be the emissive power of a perfectly black body (absorptive power $a_\lambda = 1$) at the same temperature receiving the same incident flux $E_\lambda$ from the enclosure.

The body absorbs energy $a_\lambda E_\lambda$ per unit time (per unit area, per unit wavelength interval) from the enclosure's radiation, and it emits energy $e_\lambda$. Since the body is in thermal equilibrium with the enclosure (its temperature is not changing), the energy absorbed must equal the energy emitted:

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