Free-Electron (Fermi) Energy of Tungsten: Using specific gravity 13.8, atomic weight 184.0, and two free electrons per atom, estimate the Fermi energy (in eV).

Introduction / Context:The Fermi energy E_F of a metal within the free-electron model depends on the conduction electron density n. For a given material, n is obtained from mass density and atomic weight along with the number of free electrons per atom. This problem estimates E_F for tungsten with given properties.

Given Data / Assumptions:

• Specific gravity ρ_s = 13.8 ⇒ mass density ρ = 13.8 g/cm^3 = 13,800 kg/m^3.
• Atomic weight (molar mass) M = 184.0 g/mol = 0.184 kg/mol.
• Free electrons per atom z = 2.
• Avogadro constant N_A ≈ 6.022 × 10^23 mol^−1.
• Electron mass m_e ≈ 9.11 × 10^−31 kg; Planck constant h ≈ 6.626 × 10^−34 J·s; 1 eV = 1.602 × 10^−19 J.

Concept / Approach:

Number density of atoms n_a = (ρ / M) · N_A. Conduction electron density n = z · n_a. Fermi energy is E_F = (h^2 / 2m_e) · (3π^2 n)^(2/3). Convert joules to electron-volts at the end.

Step-by-Step Solution:

Verification / Alternative check:

Cross-check with known typical Fermi energies: many transition metals lie between 6–12 eV. The computed 7.33 eV is physically reasonable and aligns with the density-scaled estimate from tungsten’s standard value.

Why Other Options Are Wrong:

• 8.95 eV: Closer to tungsten’s E_F at ρ ≈ 19.3 g/cm^3; too high for the lower density 13.8 g/cm^3 assumed here.
• Negative energies (−8.95 eV, −7.326 eV): Fermi energy is a positive energy measure above the band bottom in this model.
• 5.00 eV: Too low for a two-electron metal with high atomic density.

Common Pitfalls:

• Mistreating specific gravity units; ensure conversion from g/cm^3 to kg/m^3.
• Forgetting the 2/3 power dependence on electron density.

Final Answer:

7.326 eV