(a) (i) Using free electron theory of metals, calculate the Fermi energy level of sodium atom at absolute zero. Assume that…

(a) (i) By free electron theory at absolute zero, sodium has one conduction electron per atom. Taking atomic mass of sodium as 23 g mol⁻¹,

n = (ρ N_A)/A = (0.97 g cm⁻³ × 6.022×10²³ mol⁻¹)/(23 g mol⁻¹)

n = 2.54×10²² cm⁻³ = 2.54×10²⁸ m⁻³.

For a free electron gas,

E_F = (ħ²/2mₑ)(3π²n)^(2/3), k_F = (3π²n)^(1/3).

Thus k_F = (3π² × 2.54×10²⁸)^(1/3) = 9.09×10⁹ m⁻¹.

E_F = [(1.055×10⁻³⁴)² × (9.09×10⁹)²]/(2 × 9.11×10⁻³¹) J

E_F = 5.05×10⁻¹⁹ J = 5.05×10⁻¹⁹/(1.602×10⁻¹⁹) eV = 3.15 eV.

This holds for T = 0 K, one free electron per atom, and independent free electrons.

(a) (ii) I. Electron in a one-dimensional box For width L and infinite walls, the Schrödinger equation gives

Eₙ = n²h²/(8mL²) = n²π²ħ²/(2mL²), n = 1, 2, 3, ...

ψₙ(x) = sqrt(2/L) sin(nπx/L), 0 < x < L.

Energy level diagram:

n=3 __________ E₃ = 9E₁ n=2 __________ E₂ = 4E₁ n=1 __________ E₁ = π²ħ²/(2mL²)

II. Linear harmonic oscillator For potential V(x) = 1/2 mω²x²,

Eₙ = (n + 1/2)ħω, n = 0, 1, 2, ...

ψₙ(x) = Nₙ Hₙ(αx) exp(−α²x²/2), α = sqrt(mω/ħ).

Energy level diagram:

n=3 __________ E₃ = 7/2 ħω n=2 __________ E₂ = 5/2 ħω n=1 __________ E₁ = 3/2 ħω n=0 __________ E₀ = 1/2 ħω

Qualitative comparison: Both give discrete bound states and nonzero zero-point energy. In the box, Eₙ ∝ n², so level spacing increases; the wavefunctions are sinusoidal and vanish at the walls. In the harmonic oscillator, Eₙ is linear in n, levels are equally spaced; wavefunctions are Hermite-Gaussian and extend beyond classical turning points. The box potential is infinite and flat inside, while the oscillator potential is parabolic.

(b) Let the electronic momentum be pₑ ≈ ħ/a. Then the electronic energy scale is

E_el ≈ pₑ²/(2mₑ) ≈ ħ²/(2mₑa²).

For nuclear rotation, the two nuclei of mass M separated by distance a give a moment of inertia of order I ≈ M a². Hence the rotational energy scale is

E_rot ≈ ħ²/(2I) ≈ ħ²/(2Ma²).

Therefore

E_rot/E_el ≈ [ħ²/(2Ma²)]/[ħ²/(2mₑa²)] = mₑ/M.

Since M ≫ mₑ, E_rot is smaller than E_el by the factor mₑ/M, apart from numerical constants of order unity. By the same reasoning (Born–Oppenheimer), vibrational energies are smaller than electronic ones by about (mₑ/M)^(1/2), so E_el : E_vib : E_rot ≈ 1 : (mₑ/M)^(1/2) : mₑ/M, which is why electronic, vibrational and rotational spectra lie in the UV–visible, infrared and microwave regions respectively.

(c) In L-S coupling, electrostatic interaction between electrons dominates. The individual orbital angular momenta couple to give total L, and spins couple to give total S; then J = L + S, …, |L − S|. The good quantum numbers are L, S, J, M_J. It is common in light atoms.

In J-J coupling, spin-orbit interaction for each electron dominates. Each electron has jᵢ = lᵢ ± 1/2, and these jᵢ couple to give total J. The good quantum numbers are jᵢ, J, M_J; L and S are not well defined. It is common in heavy atoms.

For l = 1 and s = 1/2, the possible total J values are

J = l + s = 3/2 and J = l − s = 1/2.

The magnitude is |J| = sqrt(J(J + 1))ħ, and the component along a chosen z-axis is J_z = M_Jħ. The orientation angle θ satisfies

cos θ = M_J/sqrt(J(J + 1)).

For J = 3/2: M_J = 3/2, 1/2, −1/2, −3/2. sqrt(J(J + 1)) = sqrt(15)/2.

cos θ = ±3/sqrt(15) = ±sqrt(3/5) for M_J = ±3/2, θ ≈ 39.2° and 140.8°.

cos θ = ±1/sqrt(15) for M_J = ±1/2, θ ≈ 75.0° and 105.0°.

For J = 1/2: M_J = 1/2, −1/2. sqrt(J(J + 1)) = sqrt(3)/2.

cos θ = ±1/sqrt(3), θ ≈ 54.7° and 125.3°.

Thus J = 3/2 has 4 possible orientations, and J = 1/2 has 2 possible orientations; total 6 = (2l + 1)(2s + 1).