Black Body Irradiation
The practical purpose of the black body is that each body loses the energy by the emission of electromagnetic radiation. The higher temperature the object has - the radiation will be in shorter wave region: the Sun obviously send its radiation in the spectral region that corresponds to its high temperature. A hot "oven" irradiates from dark-red through ruby and yellow right up to white, depending on the higher temperature is. Even whether the oven is not so extremely hot, we can feel the irradiated light. For instance, the streaming hot water of a heater emits light in the thermal radiation form. The scientists on the border of the 19th and 20th century tried to determine the laws of the radiation spectral intensity distribution of different temperatures.

And at the same time the concept was that the light is emitted by oscillators. The oscillator has the average energy <E> = kT having the temperature T. The energetic radiation density u(ν)dν multiplied by the spectral range [ν, ν+dν] is proportional to the oscillators amount per volume unit in this range multiplied by kT:

u(ν)dν = <E> dN(ν) = kT dN(ν)

u(ν)dν = Energy of radiation in the range [ν,ν+dn]/Volume

dN(ν) was calculated by Rayleigh and Jeans: dN(ν) = 8pn²/c³ dν

u(ν) = 8pn²/c³ kT | The experiments prove the Rayleigh-Jeans law in IR region !

Since u(ν) increases proportionally to ν2 , so it leads to the "Ultraviolet Catastrophe" because there would be much more energy irradiated with the increasing frequency. The glance at the dark lecture room shows that it doesn't emit in UV, i.e. the energy distribution law is wrong (at least in UV region).

1900 Max Planck supposed that the energy are not irradiated continuously but by small fractions (or energy quanta): the electron should emit the energy quantum hν or nothing moreover h is the proportionality constant. So, it can be irradiated only integer of hν , but not for instance, 3/2hν or 1,7754hn.
Planck derived the following formula for the spectral energy density u(ν)dν = <E> · dN(ν) in the frequency range [ν, ν+dν] (Derivation):

| <E> | dN
u(ν)dν = | hν/e+hν/kT− 1 | . 8pn ²/c³ dν

u(ν) = 8πhν³/c³.1/ehν/kT− 1

This black body irradiation formula perfectly coincide with the experiments.

From the maximum of the distribution one can find νmax (differentiating upon ν and putting to zero) and the meaning of h can be determined from the experiment:

νmax = 2,8214 kT/h

h = 6,626176·10-34 Js

Integration of the Planck equation gives the radiation law of Stefan and Boltzmann:

o∫∞ u(ν)dν = U(T) = aT4 (a = 7,56·10-16 Jm-3K-4)

The intensity (radiated energy per surface and time) is I = 1/4 c · a · T4

I = σoT4[W/m²] so » 5,6697 · 10-8 [Wm-2K-4]

The radiated energy increases with the fourth power of the absolute temperature T.

Because of the scale accuracy of Planck constant h we hasn't took into account it. Below you can find two numerical examples:

1. The sand grain with the mass m = 10-6 kg is falling from the height H. How high should be the height H in order that grain's energy would be equal to the energy of the visual range quanta (νvisual ≈ 5 . 1014 Hz a Na-Lamp) ?
m · g · H = hn ® H = hν/mg = 3,38 · 10-14 m
This corresponds to about one 10-thousandth nuclear diameter!
2. How many quanta does a lamp with P = 10 W send out:
P = E/t = nhν/t ® n/t = P/hν» 3 · 1019 Quanta/s
Man can't notice if there is one quantum absent once because then the power decreases only in about 19th digit after the comma.