Fermi–Dirac Occupancy Near the Fermi Level In a metal at ordinary temperatures, how does the probability of occupancy of a state 0.1 eV below the Fermi level compare with that of a state 0.1 eV above the Fermi level?

Introduction / Context:
The Fermi–Dirac distribution gives the probability that an electronic state of energy E is occupied at temperature T: f(E) = 1 / (1 + exp[(E − EF)/(kT)]). This distribution is steep near the Fermi level EF, with occupancy dropping from near 1 below EF to near 0 above EF within a few kT. Understanding this asymmetry is essential for transport and tunneling analyses.

Given Data / Assumptions:

• Metal at moderate temperature (e.g., 300 K).
• Energy offsets ±0.1 eV relative to EF, which are several times kT (kT ≈ 0.025 eV).
• No band structure anomalies near EF.

Concept / Approach:

For ΔE = 0.1 eV, ΔE/(kT) ≈ 0.1/0.025 ≈ 4. Thus, f(EF − 0.1 eV) ≈ 1 / (1 + e^(−4)) ≈ 0.982, while f(EF + 0.1 eV) ≈ 1 / (1 + e^(+4)) ≈ 0.018. Therefore, the state below EF is far more likely to be occupied than the one above EF.

Step-by-Step Solution:

Verification / Alternative check:

Symmetry: f(EF + ΔE) = 1 − f(EF − ΔE). For ΔE ≫ kT, one is near 1, the other near 0.

Why Other Options Are Wrong:

• Equal or indeterminate contradicts the monotonic nature of f(E) around EF.
• “Less than” reverses the physically correct inequality.

Common Pitfalls:

Forgetting to convert kT to eV; assuming a symmetric 50–50 occupancy at ±ΔE (only true at ΔE = 0).

Final Answer:

Greater than the probability 0.1 eV above the Fermi level