Solid-state physics — assertion–reason on typical Fermi energy of metals Assertion (A): For most common metals, the Fermi energy E_F is less than about 10 eV. Reason (R): The Fermi level for a free-electron-like metal is approximated by E_F ≈ 3.64 × 10^−19 * n^(2/3) eV, where n is the number of free electrons per m^3.

Introduction / Context:
The Fermi energy sets the top of the occupied electron states at absolute zero in the free-electron model of metals. Typical magnitudes help estimate electron speeds, densities of states, and electronic heat capacities.

Given Data / Assumptions:

• Free-electron approximation: E_F = (h^2 / 2m_e) * (3π^2 n)^(2/3).
• Representative conduction-electron densities n ≈ 10^28–10^29 m^−3.
• Energies expressed in electron-volts (eV).

Concept / Approach:
Inserting typical n values into the free-electron formula yields E_F on the order of a few eV up to roughly 10 eV (e.g., Na ~3 eV, Al ~11.6 eV, Cu ~7 eV). The given empirical form E_F ≈ 3.64 × 10^−19 * n^(2/3) (with n in m^−3) is a convenient numerical version of the theoretical expression after unit conversion to eV.

Step-by-Step Solution:
Take n = 1 × 10^29 → n^(2/3) ≈ 2.15 × 10^19.Compute E_F ≈ 3.64 × 10^−19 * 2.15 × 10^19 ≈ 7.8 eV.Since many metals have n in this range, E_F typically lies below ~10 eV → A is true and explained by R.

Verification / Alternative check:
Using the exact formula E_F = (h^2/2m_e)(3π^2 n)^(2/3) and converting to eV reproduces the same magnitude, confirming the numerical coefficient used in R.

Why Other Options Are Wrong:
(b) denies the explanatory link even though substituting typical n into R’s formula directly supports A. (c) and (d) reject correct statements. (e) is inconsistent with established metal data.

Common Pitfalls:

• Confusing work function with Fermi energy; they are different quantities.
• Forgetting unit consistency when using empirical constants.

Final Answer:
Both A and R are true and R is correct explanation of A