According to Kirchhoff's law, the ratio of emissive power to absorptivity of all bodies is equal to the emissive power of a
black body
Kirchhoff's Law of Thermal Radiation is a fundamental principle that connects how well an object emits thermal radiation and how well it absorbs it at a specific temperature.
The law states that for any object, the ratio of its emissive power (how much radiation it emits per unit area) to its absorptivity (the fraction of incident radiation it absorbs) is equal to the emissive power of a reference object known as a black body, provided they are at the same temperature and under the same conditions.
This relationship can be shown mathematically. For a specific wavelength ($\lambda$) and temperature ($T$):
$$ \frac{E(\lambda, T)}{a(\lambda, T)} = E_b(\lambda, T) $$
Where:
This means that a surface that is a good absorber of radiation at a certain wavelength is also a good emitter of radiation at that same wavelength, compared to a black body.
A black body is an idealized concept representing a perfect absorber and emitter of thermal radiation. It absorbs all wavelengths of incident radiation completely and emits radiation based solely on its temperature, following Planck's Law.
Kirchhoff's Law essentially uses the black body as a benchmark. It tells us that the property of emitting radiation ($E$) and absorbing radiation ($a$) are closely linked, and their ratio is constant for all objects at a given temperature, equaling the emission capability of the perfect black body ($E_b$).
Stefan Boltzmann's constant is expressed in the unit-
A body whose absorptivity does not vary with temperature and wavelength of the incident ray is known as
The process of heat transfer from a hot body to a cold body in a straight line, without affecting the intervening medium, is known as ______.
Heat is transferred from an electric bulb by ______.