Rayleigh and Jeans law|Quantum|blackbodyradiation
RAYLEIGH AND JEANS LAW
Rayleigh and Jeans have derived an expression for the radiation energy per unit volume of a hot body
in the frequency range between v and v+ dv. do In Jean's derivation of radiation law it is assumed that the radiation is broken up into monochromatic wave tranins and the number of such trains or equivalent degrees of freedom lying between frequency range v and v + dv is determined. Then the energy carried by each degree of freedom is calculated from general statistical theory and hence the energy density of the radiation can be determined.
Let the radiation of wavelengtnh , enclosed between two perfectly reflecting parallel walls separated by a distance 1, form steady stationary waves. Then
l=n1.λ/2
where n1 is an integer being equal to the number of nodal planes between parallel walls.
If the wave is travelling in such a direction that the normal to the incident wavefront makes an angle e with the normal to the walls as shown in fig. , then the path difference between incident and the reflected beams at point Q
=PR+RQ
=RQ cos 2A+RQ
=(1+cos 2A) RQ
=(1+cos 2A) RO sec A
=(1+cos 2A) x secA
(since RO=x)
= 2 cos² A. x sec A
= 2x cos A. If Q lies on the nodal plane, i.e., if OQ is a nodal plane, the path difference 2x cos A must be equal to odd multiple of 1/2 i.e., for a nodal plane we must have
➤2x cos A=(2n+1)λ/2
Therefore the distance between successive nodal planes is cose and the distance of the first nodal is / c plane from the wall is λ/4cosA Therefore if we have stationary waves with ny nodal planes, between the walls separated by a distance I, then comparing with equation (1), we must have
➤n1.λ/2=I cosA....(3)
Finally let us consider a cubical box with perfectly reflecting walls of side / enclosing radiation which produces stationary waves. If the normal to the incident wave front makes angles α,β,γ with the normals to three parallel pairs of the faces of the cube and if n1, n2 and n3 are the nodal planes to first, second and third pair of face
respectively, we have
➤n1.λ/2=/cos α,
➤n2.λ/2= I cosβ
➤n3 λ/2=l cosγ
Squaring and adding, we get
➤(n1+n2+n3})²λ²/4=l² (cos²α + cos²β + cos²γ )
➤½ λ√(ni+n₂+ n3) = l......(4)
➤(since cos²α+ cos² β+ cos²α=1)
If c is the velocity of waves and v their frequency, equation (4) may be written as
➤ v=C/2l/√(n²1+n²2+n²3}) .....(5)
Now we have to calculate the number of possible stationary vibrations lying between frequency V and v+v which may be done as follows.
Let plot on three rectangular axes X, Y, Z the values n1c/2l,n2c/2l, n3c/2 respectively (Fig represents the plot in two dimensions). This gives us a simple cubic lattice of the point each having a distance equal to v from the origin. Then the total number
of n's such that
➤<C/2l/√(n²1+n²2+n²3}) <v+dv.
is the number of points lying in one octant between the spheres of radii v and v+dv. If dv is small, the volume of the spherical shell is 4nv² dv. The volume of an elementary cubic lattice corresponding to one vibration
4πν²dv/8(c/2l)³ = 4πν²l³dv/c³
Therefore the total number of independent vibrations lying between frequency v and v+dv, in the cube of volume l³.
The expression has been divided by 8 since n1, n2, n3, occur in one out of 8 octants.
Therefore the number of independent vibrations per unit volume in one out of 8 octants,
4πν²dvl²/c³l³=4πν²dv/c³......(6)
This is independent of the volume of enclosure and hence holds for a body of any shape. As the waves in ether are transverse and can be polarised; each polarised component being independent of the other, therefore the number of independent vibrations represented by (6) must be doubled, i.e.
The number of independent vibrations or degrees of freedom per unit volume with frequencies between v and v+ dv
=8πν²dv/c³.......(7)
The energy associated with each degree of freedom, according to Maxwell's Boltzmann's law
Thus according to classical theory the average energy associated with each degree of freedom is *T, so that the energy per unit volume in the frequency range vand v+ dv is given by
Eν=8πν²kT/c³dv.
This is Rayleigh-Jean's radiation law derived from classical considerations.
According to the expression the energy emitted by a black body increases with frequency and becomes in finite at large frequencies. The experimentally observed radiation curve is in complete disagreement with this conclusion since experimental curve shows that energy radiated at a particular temperature becomes zero both at lower as well as higher frequencies as shown in fig.
Moreover the energy carried by all frequencies
The value of this integral is infinity for any value of T other than zero. This physically means that the total energy emitted by a black-body per unit area per unit time is infinity at all temperatures. Clearly this is entirely false conclusion.
Both of the above conclusions are incorrect and fail completely to account for the observed spectrum of radiation emitted by a black-body. The frequency distribution of a black-body could not be explained on the basis of classical concepts in spite of repeated attempts by various scientists. The discrepany in classical calculations and observed facts was removed by Planck on the basis of quantum theory.
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