Question

In which one of the following limits does the Fermi-Dirac distribution \(n_F(\epsilon, T) = \left(e^{\frac{\epsilon - \mu}{k_B T}} + 1\right)^{-1}\) and the Bose-Einstein distribution \(n_B(\epsilon, T) = \left(e^{\frac{\epsilon - \mu}{k_B T}} - 1\right)^{-1}\) reduce to the Maxwell-Boltzmann distribution?
 

• A. \(\mu = 0\)
• B. \((\epsilon - \mu) < k_B T\)
• C. \((\epsilon - \mu) \gg k_B T\)
• D. \(\mu \gg k_B T\)

Hint:

The Maxwell-Boltzmann distribution is the classical limit of quantum statistics, valid when particle occupancy is very low.

Answer 1

The Correct Option is C

Step 1: Recall Maxwell-Boltzmann distribution form.
The Maxwell-Boltzmann (MB) distribution is given by: \[ n_{MB}(\epsilon, T) = e^{-\frac{\epsilon - \mu}{k_B T}} \]

Step 2: Approximation condition.
For both Fermi-Dirac and Bose-Einstein, if \(e^{\frac{\epsilon - \mu}{k_B T}} \gg 1\), then \[ n_F \approx e^{-\frac{\epsilon - \mu}{k_B T}}, n_B \approx e^{-\frac{\epsilon - \mu}{k_B T}} \] which corresponds to the Maxwell-Boltzmann limit.

Step 3: Conclusion.
Therefore, the MB distribution is valid when \((\epsilon - \mu) \gg k_B T.\)