# Thermal de Broglie wavelength

(Redirected from Thermal wavelength)

In physics, the thermal de Broglie wavelength ($\Lambda$) is roughly the average de Broglie wavelength of the gas particles in an ideal gas at the specified temperature. We can take the average interparticle spacing in the gas to be approximately (V/N)1/3 where V is the volume and N is the number of particles. When the thermal de Broglie wavelength is much smaller than the interparticle distance, the gas can be considered to be a classical or Maxwell–Boltzmann gas. On the other hand, when the thermal de Broglie wavelength is on the order of or larger than the interparticle distance, quantum effects will dominate and the gas must be treated as a Fermi gas or a Bose gas, depending on the nature of the gas particles. The critical temperature is the transition point between these two regimes, and at this critical temperature, the thermal wavelength will be approximately equal to the interparticle distance. That is, the quantum nature of the gas will be evident for

$\displaystyle \frac{V}{N\Lambda^3} \le 1 \ , {\rm or} \ \left( \frac{V}{N} \right)^{1/3} \le \Lambda$

i.e., when the interparticle distance is less than the thermal de Broglie wavelength; in this case the gas will obey Bose–Einstein statistics or Fermi–Dirac statistics, whichever is appropriate. This is for example the case for electrons in a typical metal at T = 300 K, where the electron gas obeys Fermi–Dirac statistics, or in a Bose–Einstein condensate. On the other hand, for

$\displaystyle \frac{V}{N\Lambda^3} \gg 1 \ , {\rm or} \ \left( \frac{V}{N} \right)^{1/3} \gg \Lambda$

i.e., when the interparticle distance is much larger than the thermal de Broglie wavelength, the gas will obey Maxwell–Boltzmann statistics.[1] Such is the case for Thermal Neutron produced by a Neutron source.

## Massive Particles

For a free ideal gas of massive particles (with no internal degrees of freedom) in equilibrium, the thermal de Broglie wavelength can be obtained through the standard de Broglie wavelength:

$\Lambda = \frac{h}{p}$, with the substitution of the momentum ($p$) by the Kinetic energy $E_K = \frac{p^2}{2m}$:
$\Lambda = \frac{h}{\sqrt{2mE_K}}$.

In the nonrelativistic case the effective Kinetic energy of free particles is $E_K=\pi k_B T$.

$\Lambda = \sqrt{\frac{ h^2}{ 2\pi mkT}} = \frac{h}{\sqrt{2\pi mkT}},$

Using the root mean square Kinetic energy of free particles gives $E_K=\frac{3}{2} k_B T$.

$\Lambda = \frac{h}{\sqrt{3 mkT}},$

where h is the Planck constant, m is the mass of a gas particle, k is the Boltzmann constant, and T is the temperature of the gas.[1]

## Massless particles

For a massless particle, the thermal wavelength may be defined as:

$\Lambda= \frac{ch}{2 \pi^{1/3} k T}$

where is the speed of light. As with the thermal wavelength for massive particles, this is of the order of the average wavelength of the particles in the gas and defines a critical point at which quantum effects begin to dominate. For example, when observing the long-wavelength spectrum of black body radiation, the "classical" Rayleigh–Jeans law can be applied, but when the observed wavelengths approach the thermal wavelength of the photons in the black body radiator, the "quantum" Planck's law must be used.

## General definition of the thermal wavelength

A general definition of the thermal wavelength for an ideal quantum gas in any number of dimensions and for a generalized relationship between energy and momentum (dispersion relationship) has been given by Yan (Yan 2000). It is of practical importance, since there are many experimental situations with different dimensionality and dispersion relationships. If n is the number of dimensions, and the relationship between energy (E) and momentum (p) is given by:

$E=ap^s\,$

where a and s are constants, then the thermal wavelength is defined as:

$\Lambda=\frac{h}{\sqrt{\pi}}\left(\frac{a}{kT}\right)^{1/s} \left[\frac{\Gamma(n/2+1)}{\Gamma(n/s+1)}\right]^{1/n}$

where Γ is the Gamma function. For example, in the usual case of massive particles in a 3-D gas we have n = 3 , and E = p2/2m which gives the above results for massive particles. For massless particles in a 3-D gas, we have n = 3 , and E = p c which gives the above results for massless particles.

## References

1. ^ a b Charles Kittel; Herbert Kroemer (1980). Thermal Physics (2 ed.). ,W. H. Freeman. p. 73. ISBN 978-0716710882.