Question

How did Plank explain the black body radiation?

Answer 1

When there is no net transfer of matter or energy between the body and its surroundings, Planck's law defines the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature $T$.

Complete answer:
The spectral-energy distribution of radiation emitted by a blackbody is explained by Max Planck's radiation law, a mathematical relationship (a hypothetical body that completely absorbs all radiant energy falling upon it, reaches some equilibrium temperature, and then re emits that energy as quickly as it absorbs it).

Planck believed that the sources of radiation are oscillating atoms, and that each oscillator's vibrational energy can take any of a number of distinct values, but never any value in between. Planck also believed that when an oscillator switches from one state of energy ${E_1}$ to ${E_2}$ put it another way, the discrete amount of energy ${E_1} - {E_2}$ , or quantum of radiation, is equal to the product of the frequency of the radiation, represented by the Greek letter, and a constant $h$ now known as Planck's constant, which he measured from blackbody radiation data; i.e. ${E_1} - {E_2} = hv$.

For the energy radiated per unit volume by a cavity of a blackbody in the wavelength interval $\lambda$ to $\lambda + \Delta \lambda$ Planck's law for energy ${E_{_\lambda }}$ may be written in terms of Planck's constant $h$, the speed of light $\left( c \right)$, the Boltzmann constant $\left( k \right)$, and the absolute temperature,
\left( T \right)$$${E_\lambda } = \dfrac{{8\pi hc}}{{{\lambda ^5}}} \times \dfrac{1}{{\exp (hc/kT\lambda ) - 1}}$$ The emitted radiation's wavelength is inversely proportional to its frequency, or \lambda = \dfrac{c}{v}$. The Planck constant's value is defined as$6.62607015 \times {10^{ - 34}} joule.\sec $

Note: The bulk of the radiation emitted by a blackbody at temperatures up to several hundred degrees is in the infrared region of the electromagnetic spectrum. At higher temperatures, the overall radiated energy rises, and the intensity peak of the emitted spectrum changes to shorter wavelengths, resulting in visible light being emitted in greater amounts.