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

[!TLDR] Kirchhoff's law: a body's emissivity equals its absorptivity (a=e); proved via thermal equilibrium between an ordinary body and a blackbody in an enclosure. [!ANSWER] Kirchhoff's law: a = e for every body, at every wavelength, in thermal equilibrium -- prove …

STATEMENT: Kirchhoff's law of thermal radiation states that, at a given temperature, the ratio of a body's emissive power to its coefficient of absorption equals the emissive power of a perfect blackbody at that same temperature, for every wavelength. Equivalently, for a body emitting and absorbing radiation in thermal equilibrium, its emissivity equals its absorptivity: a = e (or a(λ) = e(λ), wavelength by wavelength).

PROOF: Consider an ordinary body A and a perfect blackbody B, of identical geometric shape, placed together inside a common enclosure. Once thermal equilibrium is reached, both bodies (and the enclosure) share the same temperature. Let R be the emissive power of A, RB the emissive power of B, and a the coefficient of absorption of A. If Q is the radiant heat incident on each body per unit time, the heat absorbed by A is Qa = aQ. Since A's temperature stays constant, it must emit exactly as much as it absorbs per unit time, so aQ = R. For the blackbody B, since a = 1 (it absorbs everything), the same equilibrium condition gives Q = RB. Dividing the first equation by the second: a = R/RB. But by the definition of emissivity, e = R/RB. Therefore a = …

Set up the thermal-equilibrium condition (emission rate = absorption rate) separately for an ordinary body and for a blackbody sharing an encl …

Forgetting to use a = 1 for the reference blackbody B when writing its own equilibrium condition (Q = RB), which is the key simplif …