State Kirchhoff's law of radiation and give its theoretical proof.
Answers
For a body of any arbitrary material emitting and absorbing thermal electromagnetic radiation at every wavelength in thermodynamic equilibrium, the ratio of its emissive power to its dimensionless coefficient of absorption is equal to a universal function only of radiative wavelength and temperature. That universal function describes the perfect black-body emissive power.[1][2][3][4][5][6]
Here, the dimensionless coefficient of absorption (or the absorptivity) is the fraction of incident light (power) that is absorbed by the body when it is radiating and absorbing in thermodynamic equilibrium.
In slightly different terms, the emissive power of an arbitrary opaque body of fixed size and shape at a definite temperature can be described by a dimensionless ratio, sometimes called the emissivity: the ratio of the emissive power of the body to the emissive power of a black body of the same size and shape at the same fixed temperature. With this definition, Kirchhoff's law states, in simpler language:
For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.
In some cases, emissive power and absorptivity may be defined to depend on angle, as described below. The condition of thermodynamic equilibrium is necessary in the statement, because the equality of emissivity and absorptivity often does not hold when the material of the body is not in thermodynamic equilibrium.
Kirchhoff's law has another corollary: the emissivity cannot exceed one (because the absorptivity cannot, by conservation of energy), so it is not possible to thermally radiate more energy than a black body, at equilibrium. In negative luminescence the angle and wavelength integrated absorption exceeds the material's emission, however, such systems are powered by an external source and are therefore not in thermodynamic equilibrium.
Answer:
Explanation:
Kirchhoff's law of radiation : At a given temperature, the coefficient of absorption of a body is equal to its coefficient of emissoin.
Theoretical proof : Kirchhoff's law can be proved with the help of the following thought experiment. Suppose that an ordinary body A and blackbdoy B are enclosed in an athermanous enclosure. According to Prevost's theory of heat exchanges, there is a continuous exchange of heat between each body and its surroundings Hence, the two bodies, after some time, will attain the same temperature as that of the enclosure. Let a and e be respectively the coefficients of absorption and emission of body A. Let E be the emissive powers of bodies A and B, respectively. Suppose that Q is the quantity of radiant energy incident on each body per unit time per unit surface area of the body. Body A will absorb the quantity aQ per unit time per unit surface area and radiate the quantity E per unit time per unit surface area. Since there is no change in its temperature
As body B is a blackbody, it will absorb the quantity Q per unit time per unit surface area and radiate the quantity
per unit time per unit surface area. Since there is no change in its temperature.
From Eqs. (1) and (2), we get,
........(3) <br> By definition of coefficient of emission, <br>
...........(4) <br> From Eqs. (3) and (4), we get, a=e <br> Hence, the proof of Kirchhoff's law.