The Black Body Radiator

A black body radiator loses energy by emitting electromagnetic radiation. Because temperature has an inverse relationship with wavelength, when the temperature of the black body raises, the wavelength of the radiation that it produces gets shorter. The Sun emits radiation in the visible region of the spectrum which indicates that its has a high temperature.  A hot oven radiates in the dark-red through ruby and yellow through white regions of the spectrum depending on how hot it is. Even when the oven is not so hot, we feel the irradiated light waves. An example is the flowing hot water of a heater, which emits electromagnetic radiation in the form of thermal radiation. Scientists in the 19th and 20th century tried to determine the laws of the spectral intensity distribution at different temperatures. 

The prevalent theory at the time was that light was produced by miniature oscillating bodies.  In this theory, these oscillators have an average energy of <E> = kT at a temperature T. The radiation's energy density u(ν)dν  multiplied in the spectral range [ν, ν+dν]  is proportional to the number of osciallations per volume unit 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 , an "Ultraviolet Catastrophe" results because there would be an unreasonable amount of energy radiated with a high frequency. 

1900 Max Planck supposed that a photon cannot have any value of energy but instead only discrete values (or quanta): An electron should have one quanta of energy since it is the most fundamental particle. The energy of an electromagnetic wave is described by E=nhν where, h is a proportionality constant. It follows that only electromagnetic radiation with an integer value for n in the above equation can be produced.
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
.  8pn²/c³ dν
u(ν)  =  8πhν³/c³.1/ehν/kT− 1

This black body radiation formula perfectly coincides with the experiments.

From the maximum of the distribution one can f 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.

Be aware that we havent yet ensured the scale accuracy of Planck constant h. Below are two problems involving the above concepts:

  1. A grain of sand with a mass m = 10-6 kg falls from a height H. How high should the height H be for the grain's energy before impact to 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 because then the power decreases by only an infinitessimal amount.




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