(a) Show that for an n-type semiconductor, the Fermi level lies midway between the donor states and the conduction band edge at…

(a) Fermi level of an n-type semiconductor at low temperature

Model: energies are measured from the valence band top (E_v = 0), so E_c = E_g. Donor levels E_D lie a small ionisation energy E_d below E_c (E_D = E_c − E_d), with N_D donors per unit volume. At low temperature kT is much less than E_g, so holes are negligible (p ≈ 0), and only a few donors are ionised, so E_F lies between E_D and E_c with E_c − E_F ≫ kT (Boltzmann statistics apply).

Electrons in the conduction band: n = N_c exp[−(E_c − E_F)/kT], with N_c = 2(2πm_e*kT/h²)^(3/2).

Ionised donors: a donor holds at most one electron (two spin orientations), so N_D⁺ = N_D/[1 + 2 exp((E_F − E_D)/kT)] ≈ (N_D/2) exp[−(E_F − E_D)/kT], since E_F − E_D ≫ kT.

Charge neutrality (every conduction electron comes from a donor): n = N_D⁺, N_c exp[−(E_c − E_F)/kT] = (N_D/2) exp[−(E_F − E_D)/kT].

Taking logarithms: (E_F − E_D) − (E_c − E_F) = kT ln(N_D/2N_c), so E_F = (E_c + E_D)/2 + (kT/2) ln[N_D/(2N_c)], or with E_v = 0, E_F = E_g − E_d/2 + (kT/2) ln(N_D/2N_c).

As T → 0 the second term vanishes (kT ln T → 0), so E_F → (E_c + E_D)/2: the Fermi level lies midway between the donor level and the conduction band edge.

Consequence: n = (N_c N_D/2)^(1/2) exp(−E_d/2kT), the freeze-out law with activation energy E_d/2, analogous to E_g/2 for intrinsic material. As T rises the donors ionise fully (n ≈ N_D), E_F = E_c − kT ln(N_c/N_D) drops below E_D, and at high T it moves towards mid-gap (intrinsic behaviour).

(b) Avogadro number from X-ray diffraction

Data: λ = 0.1512 nm (the printed 0.1512 m must be 0.1512 nm, a Cu X-ray line), (111) plane, θ = 20.2°, ρ = 2698 kg m⁻³, M = 26.98 kg kmol⁻¹, fcc so 4 atoms per cell, first order.

1. Bragg’s law: d₁₁₁ = λ/(2 sin θ) = 0.1512 × 10⁻⁹/(2 × 0.34530) = 2.1894 × 10⁻¹⁰ m.
2. Lattice constant: d₁₁₁ = a/√3 (all-odd indices, allowed for fcc), a = 3.7922 × 10⁻¹⁰ m, a³ = 5.4534 × 10⁻²⁹ m³.
3. Density: ρa³ = nM/N_A, so N_A = 4 × 26.98/(2698 × 5.4534 × 10⁻²⁹) = 7.33 × 10²⁶ kmol⁻¹ = 7.33 × 10²³ mol⁻¹.

Units: (kg kmol⁻¹)/(kg m⁻³ × m³) = kmol⁻¹.

Comment: this is about 22% above the accepted 6.022 × 10²³ mol⁻¹. The crystal (ρ = 2698 kg m⁻³, M = 26.98) is aluminium, with a ≈ 0.405 nm, for which the (111) reflection of 0.1512 nm X-rays would occur at about 18.9°, not 20.2°. The method is the required one, the discrepancy lies in the printed data.

(c) Allowed and forbidden decays

Check charge Q, baryon number B, lepton numbers L_e and L_μ, angular momentum and energy.

(i) n → p + e⁻ + ν̄ₑ: allowed. Q 0 → 0, B 1 → 1, L_e 0 → +1 − 1 = 0, spin ½ → three spin-½ particles possible, Q-value 0.78 MeV. Quark flavour changes (d → u) with neutrino emission, so it is the weak interaction (β-decay).

(ii) Λ⁰ → π⁺ + π⁻: forbidden. Q is conserved, but B changes from 1 to 0, violating baryon-number conservation. It also violates angular momentum: spin ½ cannot become two spin-0 pions with integral orbital angular momentum. (Λ⁰ actually decays weakly to p + π⁻.)

(iii) π⁻ → e⁻ + γ: forbidden. L_e changes from 0 to 1, violating lepton-number conservation, and a spin-0 pion cannot give a spin-½ electron plus a photon (half-integral final J), violating angular momentum conservation.

(iv) π⁰ → e⁻ + e⁺ + νₑ + ν̄ₑ: allowed. Q, B and L_e (+1 − 1 + 1 − 1 = 0) are conserved, four spin-½ particles can couple to J = 0, and 135 MeV is far above threshold. Neutrinos can only come from the weak interaction, so the branch is extremely rare compared with π⁰ → γγ (electromagnetic).

(v) π⁺ → e⁻ + e⁺ + μ⁺ + ν_μ: allowed. Q +1 → +1, L_e 0 → 0, L_μ 0 → −1 + 1 = 0, B = 0, J = 0 possible, and 139.6 MeV exceeds 105.7 + 1.02 MeV. It is the weak decay π⁺ → μ⁺ν_μ with an extra e⁺e⁻ pair from an internal photon (electromagnetic vertex), a rare branch.