(a) How could you establish that νₑ and ν̄ₑ are two different particles ? 10 marks (b) What is the age of a fossil that contains…

(a) The electron neutrino νₑ and electron antineutrino ν̄ₑ are established to be two distinct particles by lepton-number conservation and by their opposite helicities and different weak reactions.

• Lepton number: assign Lₑ(νₑ)=+1 and Lₑ(ν̄ₑ)=−1. In all observed weak processes, total electron lepton number is conserved. Therefore a reaction that produces νₑ cannot be replaced by one producing ν̄ₑ without violating Lₑ conservation.
• Helicity: in the massless limit, νₑ is left-handed, i.e. spin opposite to momentum, while ν̄ₑ is right-handed. The charged weak current couples to left-handed neutrinos and right-handed antineutrinos.
• Charge-current reactions distinguish them. νₑ can initiate νₑ + n → p + e⁻, while ν̄ₑ can initiate ν̄ₑ + p → n + e⁺. The Reines–Cowan experiment detected reactor ν̄ₑ through the second reaction by observing positron–neutron coincidences. Solar νₑ are detected by radiochemical experiments such as νₑ + ³⁷Cl → ³⁷Ar + e⁻.
• If νₑ and ν̄ₑ were the same particle, both inverse-beta reactions would be allowed for the same particle and lepton number would not be conserved. Thus νₑ and ν̄ₑ are distinct particle and antiparticle.

(b) Use the radioactive-decay law A = A₀ exp(−λt), with λ = ln2 / T₁/₂. The half-life is T₁/₂ = 5730 yr = 5730 × 3.156 × 10⁷ s = 1.808 × 10¹¹ s. Hence λ = 0.693 / (1.808 × 10¹¹) = 3.835 × 10⁻¹² s⁻¹.

For 6 g of carbon, the number of ¹²C atoms is N(¹²C) = (6 / 12) × 6.022 × 10²³ = 3.011 × 10²³. Using the printed ratio ¹⁴C/¹²C = 1.3 × 10⁻¹³, N₀(¹⁴C) = 3.011 × 10²³ × 1.3 × 10⁻¹³ = 3.914 × 10¹⁰. Initial activity: A₀ = λN₀ = 3.835 × 10⁻¹² × 3.914 × 10¹⁰ = 0.150 s⁻¹ = 9.0 decays/min. But the stated current activity is 27 decays/min, which is larger than this A₀. The current number of ¹⁴C atoms implied by 27 decays/min is N = A/λ = (27/60) / (3.835 × 10⁻¹²) = 1.173 × 10¹¹. This is larger than the initial number 3.914 × 10¹⁰ computed from the printed ratio. Therefore the printed ratio 1.3 × 10⁻¹³ is inconsistent with the other data for a 6 g carbon sample; no physically positive age follows from it.

If the standard living ratio ¹⁴C/¹²C = 1.3 × 10⁻¹² is intended, then N₀ = 3.914 × 10¹¹, so A₀ = 90 decays/min. Then t = (1/λ) ln(A₀/A) = (T₁/₂ / ln2) ln(A₀/A) = (5730 / 0.693) ln(90 / 27) = 8268 × ln(3.333) = 8268 × 1.204 = 9.95 × 10³ yr. Age ≈ 9.95 × 10³ years using the ratio required for consistency with the measured activity. This assumes a closed system, no contamination, and a constant initial ¹⁴C/¹²C ratio.

(c) (i) K⁺ → Π⁺ + Π⁺ + Π⁻ : weak interaction, since strangeness changes, ΔS = 1. (ii) Π⁺ + p → Π⁺ + Π⁺ + n : strong interaction, no strangeness change. (iii) Π⁺ + p → Δ⁺⁺ → Π⁺ + p : strong interaction, through the Δ⁺⁺ resonance. (iv) Σ⁰ → Λ⁰ + γ : electromagnetic interaction. (v) Σ⁺ → Λ⁰ + e⁺ + νₑ : weak interaction, semileptonic charged-current decay, with ΔS = 0. (vi) K⁻ + p → K⁺ + K⁰ + Ω⁻ : strong interaction, associated production with strangeness conserved. (vii) Π⁰ → γ + e⁺ + e⁻ : electromagnetic interaction, Dalitz decay of Π⁰. (viii) Σ⁻ → n + e⁻ + ν̄ₑ : weak interaction, ΔS = 1 semileptonic decay. (ix) Λ⁰ → p + e⁻ + ν̄ₑ : weak interaction, ΔS = 1 semileptonic decay. (x) e⁺ + e⁻ → γ + γ : electromagnetic interaction, electron–positron annihilation.

(d) Let the direct lattice be generated by a, b, c. The reciprocal lattice is defined by a* = 2π (b × c) / [a · (b × c)], b* = 2π (c × a) / [a · (b × c)], c* = 2π (a × b) / [a · (b × c)]. Thus a · a* = 2π, b · b* = 2π, c · c* = 2π, while a · b* = a · c* = 0, etc. Any reciprocal lattice vector is G = h a* + m b* + n c*, where h, m, n are integers.

Let the incident wave vector be k₀ and the scattered wave vector be k₁. For elastic scattering, |k₀| = |k₁| = 2π / λ. The phase difference between waves scattered from two lattice points separated by R = n₁a + n₂b + n₃c is φ = (k₁ − k₀) · R. For constructive interference from the entire lattice, φ must be 2π times an integer for every lattice vector R. Therefore (k₁ − k₀) · R = 2π × integer. Taking R = a, b, c separately gives the Laue equations: (k₁ − k₀) · a = 2πh, (k₁ − k₀) · b = 2πm, (k₁ − k₀) · c = 2πn. These are satisfied precisely when k₁ − k₀ = h a* + m b* + n c* = G. Hence the diffraction condition in reciprocal-lattice form is k₁ = k₀ + G. This is the Laue diffraction condition.

For the Bragg form, let the incident and scattered beams make angle θ with a set of lattice planes. Then |G| = |k₁ − k₀| = 2|k₀| sinθ = 4π sinθ / λ. For planes of spacing d, the reciprocal lattice vector magnitude is |G| = 2πN / d, where N is the order. Equating gives 4π sinθ / λ = 2πN / d, so 2d sinθ = Nλ. Thus the conditions are known as the Laue diffraction conditions; the equivalent angle form is Bragg’s law.

(e) For a semiconductor, the electron and hole concentrations are n = N_c exp[−(E_c − E_F) / k_B T], p = N_v exp[−(E_F − E_v) / k_B T]. The intrinsic carrier concentration nᵢ satisfies nᵢ = N_c exp[−(E_c − E_i) / k_B T] = N_v exp[−(E_i − E_v) / k_B T], where E_i is the intrinsic Fermi level. Therefore n = nᵢ exp[(E_F − E_i) / k_B T], p = nᵢ exp[(E_i − E_F) / k_B T], and the mass-action law gives np = nᵢ².

For an n-type semiconductor, donor impurities supply conduction electrons. When donors are fully ionized and N_D ≫ nᵢ, charge neutrality gives approximately n ≈ N_D. Thus n > nᵢ. From n / nᵢ = exp[(E_F − E_i) / k_B T] > 1, we get E_F > E_i. Therefore the Fermi level shifts upward from the intrinsic level toward the conduction band. Equivalently, E_c − E_F = k_B T ln(N_c / N_D). Since N_D > nᵢ, this separation is smaller than the intrinsic separation E_c − E_i, so E_F lies closer to E_c.

For a p-type semiconductor, acceptor impurities create holes. When acceptors are fully ionized and N_A ≫ nᵢ, p ≈ N_A > nᵢ. From p / nᵢ = exp[(E_i − E_F) / k_B T] > 1, we get E_i > E_F. Thus the Fermi level shifts downward from the intrinsic level toward the valence band. Equivalently, E_F − E_v = k_B T ln(N_v / N_A). Since N_A > nᵢ, this separation is smaller than the intrinsic separation E_i − E_v, so E_F lies closer to E_v. In heavily doped cases, E_F can even enter the conduction or valence band.