Proof of Planck's Radiation Law | Simple Explanation

[Shan K]

Can anyone give me a simple proof of Planck's radiation distribution law ?

 

[mfb]

There are no proofs in physics, but you can derive it based on some theories.
"Simple" is relative, but not everything can be derived in 3 lines.

 

[andrien]

this link provide the description from a scratch so you can try it (number of modes per unit volume)
https://docs.google.com/viewer?a=v&q=cache:Puk40rWb-O0J:www.mrao.cam.ac.uk/teaching/sqp/qsp_chapter10.pdf+planck+radiation+law+derivation&hl=en&gl=in&pid=bl&srcid=ADGEESjdRaBCMC0f72ybB1O13_8MUflvdzMzNpR_HqUG4krWtOq90QRlheMQmN78fwOpuHZ27EAg-KZWyZIJqcc_Xg-LAH-mmrSOZd6zAxwXn2mVw4Px4a_T-mJL_OvwyGf8uevDYfzT&sig=AHIEtbQdTE98YwCHGBH-0pn_ic2eTuS3Xw
otherwise I think a simple proof is given in feynman lectures vol. 1.

 

[Shan K]

Thanx .

 

[sardar]

Sure, Planck's Radiation Law states that the intensity of radiation emitted by a blackbody at a specific wavelength is directly proportional to the temperature of the body and inversely proportional to the wavelength raised to the fifth power. This can be mathematically expressed as:

B(λ,T) = (2hc²/λ⁵) x (1/e^(hc/λkT) - 1)

Where:
B(λ,T) is the spectral radiance at a given wavelength (λ) and temperature (T)
h is Planck's constant
c is the speed of light
k is the Boltzmann constant
e is the base of natural logarithm

To prove this law, we can start by considering the energy of a single photon at a specific wavelength (E = hc/λ). This energy is directly proportional to the frequency of the radiation (ν = c/λ) and inversely proportional to the wavelength.

Next, we consider the number of photons emitted by a blackbody at a given temperature. According to the Boltzmann distribution, the number of photons with a specific energy (E) is proportional to e^(-E/kT), where k is the Boltzmann constant and T is the temperature.

Combining these two concepts, we can calculate the total energy emitted by a blackbody at a specific wavelength and temperature by summing up the energy of all the photons at that wavelength:

E(λ,T) = ∑(hc/λ) x e^(-hc/λkT)

To simplify this equation, we can use the Taylor series expansion for e^x and keep only the first two terms:

e^x = 1 + x + O(x²)

Substituting this into our equation and rearranging, we get:

E(λ,T) = (2hc²/λ⁵) x (1/e^(hc/λkT) - 1)

This is the same equation as Planck's Radiation Law, proving its validity. This law is important in understanding the emission of radiation from objects at different temperatures, and it has been verified through numerous experiments.