Fermi level occupancy at absolute zero If WF denotes the Fermi energy in a metal at T = 0 K, which statement correctly describes the occupation of energy levels at and around WF?

Introduction / Context:
The Fermi–Dirac distribution governs how electrons occupy energy states in metals and semiconductors. At absolute zero, thermal excitations vanish, and the distribution becomes a step function—central to solid-state physics and transport theory.

Given Data / Assumptions:

• Temperature T = 0 K (absolute zero).
• Non-interacting electron gas picture in a metal (free-electron or nearly-free-electron model).
• WF is the Fermi energy.

Concept / Approach:

The Fermi–Dirac occupation is f(E) = 1 / (1 + exp[(E − WF) / (kT)]). As T → 0, for E < WF the exponential → 0 and f → 1; for E > WF the exponential → ∞ and f → 0. Thus, all states below WF are fully occupied and all above are empty. States exactly at WF are half-occupied in an idealized continuous density of states, but in a metal this corresponds to the surface separating filled and empty states.

Step-by-Step Solution:

Verification / Alternative check:

Metals conduct because the Fermi level intersects available bands; at T = 0, electrons at WF can be excited by fields into nearby empty states above WF, enabling conduction when a field is applied.

Why Other Options Are Wrong:

• Above filled/below empty: reverses the step—physically incorrect.
• “Some below and above filled” or “both sides filled”: contradicts the step at T = 0.
• “Half at every energy”: describes neither Fermi–Dirac behavior nor any physical limit.

Common Pitfalls:

Confusing the T = 0 step function with finite-temperature smearing; at room temperature, f(E) near WF is slightly less sharp but still close to a step.

Final Answer:

All energy levels below WF are filled; all those above WF are empty