Question:medium

Quantum statistics changes into classical statistics if: (Symbols have their usual meaning)

Show Hint

The classical limit is the "non-crowded" limit. If there are many more available quantum "seats" (\(g_i\)) than there are particles (\(n_i\)) to sit in them, the particles are unlikely to interact in a way that requires quantum rules, so classical statistics work fine.
Updated On: Feb 10, 2026
  • \(\frac{g_i}{n_i} = 1\)
  • \(\frac{g_i}{n_i} \gg 1\)
  • \(\frac{g_i}{n_i} \ll 1\)
  • \(\frac{g_i}{n_i} = 0\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Define terms. - \(n_i\): Number of particles at energy level \(i\). - \(g_i\): Degeneracy of energy level \(i\), i.e., number of available states. - \(\frac{n_i}{g_i}\): Occupation index (average particles per state).
Step 2: Differentiate quantum and classical statistics. - Quantum Statistics (Fermi-Dirac, Bose-Einstein): Applicable when the occupation index is not small. Particle distinguishability and occupation limits (Pauli exclusion principle for fermions) are crucial, indicating high particle density. - Classical Statistics (Maxwell-Boltzmann): Valid for a small occupation index. Particles are treated as distinguishable with no occupation limits, applicable when particles are sparsely distributed across many states.
Step 3: State the condition for the classical limit. The classical limit is reached when the occupation index is much less than 1: \[ \frac{n_i}{g_i} \ll 1 \] This means the number of available states \(g_i\) significantly exceeds the number of particles \(n_i\). \[ g_i \gg n_i \implies \frac{g_i}{n_i} \gg 1 \] This scenario describes a low-density, high-temperature gas where quantum effects are negligible.
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