Question

2.1 de Broglie wavelengths of particles, “quantum” behavior in a gas

Consider a gas of atoms, each of mass m, confined in a container at a temperature T. Assume that the density of atoms in this gas is ρ. Using the law of equipartition of energy, derive an expression for the rms velocity v in terms of the temperature. Use this result to calculate the average de Broglie wavelength λ of atoms in this gas. At what temperature TQ does the average interparticle spacing become comparable to the de Broglie wavelength? Is there any special physical significance to this temperature? For a gas of ⁸⁷Rb (rubidium) atoms at a density of 10¹² atoms/cm³ , calculate TQ

Answer 1

The Equipartition theorem states that molecules in thermal equilibrium have the same average energy associated with each independent degree of freedom of their motion., where T is the absolute temperature.

In an ideal gas in three dimensions, there are 3 degree of freedom and so the average energy is

In Ideal gas all energy is due to the kinetic energy. So we have

, which is the rms velocity.

Now de-Broglie wavelength is given by

On substituting the value of RMS velocity we got, the kinetic energy becomes . Substituting this we have

Interparticle distance is given by , where n is the number density of the gas.

At temp T_Q, thermal de-broglie wavelength and Interparticle distance become comparable. so

At this temperature, the gas can no longer be considered a classical gas and quantum statistics has to be applied.

For ⁸⁷Rb, mass m = 1.44 x 10^-25 kg