Question

What is the black body radiation. I didn't understand it clearly

Answer 1

An object at any temperature will emit thermal radiation from its surface and the radiation emitted is dependent on the temperature of the object as well as its surface characteristics.

• At room temperature, emitted radiation: Infrared

A blackbody is a theoretical object that is able to absorb ALL incident radiation, hence appearing perfectly black since it is supposed to reflect nothing.

• Will still emit radiation characteristic of its temperature
• The radiation is known as blackbody radiation or cavity radiation
• Can be approximated using a small hole in a larger cavity. Any radiation entering the hole would have to reflect off the walls of the cavity multiple times before it escapes and is almost certain to be absorbed by the walls, regardless of what the walls are made of or the wavelength of the radiation. When the cavity is heated, the radiation emitted is continuous, and does not depend on the material in the cavity but only the temperature of the cavity walls.

Two Important Experimental Findings Regarding Blackbody Radiation:

1. The total power of the emitted radiation increases with temperature

• Stefan’s Law: P=σAeT4

• P is the power of radiation emitted from surface of object, σ is the Stefan-Boltzmann constant (5.670 x 10^-8 W m ^-2 K ^-4), A is the surface area of the object, e is the emissivity of the surface and T is surface temperature.

2. The peak of the wavelength distribution shifts to the shorter wavelengths as temperature increases

• Wien’s displacement law: λmax=bT

• λ_max is the peak wavelength, T is the absolute temperature of the blackbody, and b is a constant of proportionality called Wien’s displacement constant, equal to 2.898×10⁻³ mK

Hope this helps you