Fermi–Dirac statistics at the Fermi level For an electron energy level exactly equal to the Fermi level EF, what is the Fermi–Dirac occupation probability f(EF) at thermal equilibrium?

Introduction / Context:Fermi–Dirac statistics govern the occupancy of electron energy states in metals and semiconductors. The Fermi level EF is a key reference energy at which the probability of occupation yields an elegant and temperature-independent result. This concept underpins carrier distributions, density-of-states calculations, and device physics.

Given Data / Assumptions:

• Thermal equilibrium and non-degenerate statistics are not assumed; the exact Fermi–Dirac form is used.
• We evaluate the probability precisely at E = EF.
• Temperature T can be any non-negative absolute temperature.

Concept / Approach:

The Fermi–Dirac distribution is f(E) = 1 / (1 + exp[(E − EF) / (kT)]). At E = EF, the exponential term becomes exp(0) = 1, hence f(EF) = 1 / (1 + 1) = 1/2. This holds at all temperatures, including T = 0 K and T > 0 K, making 0.5 a universal result at the Fermi level in equilibrium.

Step-by-Step Solution:

Verification / Alternative check:

At T = 0 K, all states below EF are filled and above EF are empty; the limiting value at EF is 0.5 by convention. At finite T, thermal smearing occurs, but the symmetry about EF keeps f(EF) = 0.5.

Why Other Options Are Wrong:

0 or 1 correspond to energies far from EF at T = 0 K; 0.25 is not supported by the Fermi–Dirac formula at E = EF; “Depends on temperature” is incorrect because the value remains 0.5 for any T in equilibrium.

Common Pitfalls:

Confusing f(EF) with the average occupancy near EF where temperature does influence the slope but not the exact mid-point probability at E = EF.

Final Answer:

0.5