Question

A simple model for conduction particles in a metal predicts that

σ = (e²nl)/(2mu)

where σ is the conductivity, n is the density of particles, u is the typical velocity of random motion of the particle, m is the particle mass and l is the typical distance the particle travels between collisions.

Assume here that the particles are Bosons instead of electrons (Fermions). Obtain an expression for σ, in terms of the particle mass and the gas volume and eliminating u, if the particles are described by an ideal Boson gas at zero temperature.

1) Describe how you would proceed (making reference to the textbook) to obtain the result.

2) In which case, Boson or Fermion, would you expect the conductivity to be greater, all other things being equal? Explain.

Answer 1

1)

If is the mobility of the conducting particles inside a conductor and l is their mean free path then we know the conductivity of the particles can be defined as   

Where n is the number of particles per unit volume inside the conductor.

We have here the velocity of random motion of the particles are u

Then the change of momentum for a collision = mu-(-mu)=2mu

Hence we have the change of momentum per unit length within the distance the particle travel between collision is

Hence we have the mobility of the particles

Here in this derivation we have taken into consideration that all the particles are identical that is the particles are bosons.

Hence we have the conductivity of the particles

Hence we have

b)

In case of Fermions the particles are distinguishable that is different particles have different mass, different number density and also corresponding mean free path. Since there will be a certainty of particle mass higher than average mass hence the corresponding mean free path will also be low for higher mass particles. Which implies the Fermions will have lower conductivity than that of Bosons.

Hence the Bosons will have higher conductivity.