(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) Showing that νₑ and ν̄ₑ are different particles

Distinct particles are shown by reactions that one can induce and the other cannot. Assign electron-lepton number Lₑ = +1 to νₑ and −1 to ν̄ₑ; it is conserved in all weak processes. Antineutrinos from a reactor (from β⁻ decay, n → p + e⁻ + ν̄ₑ) cause ν̄ₑ + p → n + e⁺, which conserves Lₑ (Reines and Cowan, 1956, using positron-neutron coincidences). The same antineutrinos do not cause ν̄ₑ + ³⁷Cl → ³⁷Ar + e⁻, which would change Lₑ by 2 (Davis, 1955, found no ³⁷Ar above the background). In contrast, neutrinos from the Sun, which arise in β⁺ and electron-capture processes, do cause νₑ + ³⁷Cl → ³⁷Ar + e⁻ (the Homestake experiment). So the two beams behave differently on the same target, and the difference is explained by lepton-number conservation. Their helicities also differ: νₑ is left-handed and ν̄ₑ right-handed (Goldhaber, 1958, for νₑ). This establishes the distinction as lepton-number conserving, though it does not by itself exclude a small Majorana mass.

(b) Age of the fossil by carbon dating

Decay constant: T₁/₂ = 5730 yr = 1.808×10¹¹ s, so λ = 0.693/1.808×10¹¹ = 3.83×10⁻¹² s⁻¹. Atoms of ¹²C in 6 g: N = (6/12)(6.022×10²³) = 3.01×10²³. With the printed ratio 1.3×10⁻¹³, N₀(¹⁴C) = 3.9×10¹⁰ and A₀ = λN₀ = 0.15 s⁻¹ = 9 decays/min. This is less than the observed 27 decays/min, so the printed ratio cannot describe a living sample (it would give a negative age). Using the standard living-tissue ratio ¹⁴C/¹²C = 1.3×10⁻¹²: N₀ = 3.9×10¹¹, A₀ = 0.15 × 10 = 1.50 s⁻¹ = 90 decays/min. From A = A₀e^(−λt): t = (T₁/₂/ln 2) ln(A₀/A) = (5730/0.693) ln(90/27) = 8267 × 1.204 ≈ 9.95×10³ years. Assumptions: constant atmospheric ¹⁴C/¹²C, closed system, no contamination.

(c) Interactions in the ten reactions

(i) K⁺ → π⁺π⁺π⁻: weak (strangeness changes by 1). (ii) π⁺ + p → π⁺π⁺n: strong (S, B, Q conserved). (iii) π⁺ + p → Δ⁺⁺ → π⁺ + p: strong (resonance, lifetime about 10⁻²³ s). (iv) Σ⁰ → Λ⁰ + γ: electromagnetic (photon emitted, S conserved). (v) Σ⁺ → Λ⁰ e⁺ νₑ: weak (leptons emitted). (vi) K⁻ + p → K⁺K⁰Ω⁻: strong (strangeness −1 before and after). (vii) π⁰ → γ e⁺ e⁻: electromagnetic (Dalitz decay). (viii) Σ⁻ → n e⁻ ν̄ₑ: weak (ΔS = 1, leptons emitted). (ix) Λ⁰ → p e⁻ ν̄ₑ: weak (ΔS = 1, leptons emitted). (x) e⁺e⁻ → γγ: electromagnetic (annihilation).

(d) Diffraction conditions from the reciprocal lattice

For direct primitive vectors a, b, c, the reciprocal vectors are a* = 2π(b × c)/V, b* = 2π(c × a)/V, c* = 2π(a × b)/V, with V = a·(b × c), so that a·a* = 2π and a·b* = 0, etc. Reciprocal lattice vectors are G = ha* + kb* + lc* (h, k, l integers). Let k₀ and k be the incident and scattered wave vectors, with |k| = |k₀| = 2π/λ for elastic scattering. Waves scattered from lattice points separated by R differ in phase by (k − k₀)·R. Constructive interference from every lattice point needs this to be 2π × integer for all R = n₁a + n₂b + n₃c, so (k − k₀)·a = 2πh, (k − k₀)·b = 2πk, (k − k₀)·c = 2πl. These are satisfied together exactly when k − k₀ = G, that is k = k₀ + G (equivalently Δk = G). These are the Laue conditions (Laue equations). Bragg form: squaring k₀ + G with |k| = |k₀| gives 2k₀·G + G² = 0. With θ the glancing angle to the planes of spacing d, |G| = 2πn/d and 2|k₀| sinθ = |G|, so 2d sinθ = nλ, Bragg's law. Geometrically, diffraction occurs when the Ewald sphere (radius 2π/λ) passes through a reciprocal lattice point.

(e) Position of the Fermi level in doped semiconductors

For a non-degenerate semiconductor, n = N_c exp[−(E_c − E_F)/kT] and p = N_v exp[−(E_F − E_v)/kT]. With the intrinsic level E_i (where n = p = nᵢ): n = nᵢ exp[(E_F − E_i)/kT], p = nᵢ exp[(E_i − E_F)/kT], and np = nᵢ². n-type: donors N_D are ionised, and charge neutrality n = p + N_D gives n ≈ N_D ≫ nᵢ. Then exp[(E_F − E_i)/kT] = N_D/nᵢ > 1, so E_F − E_i = kT ln(N_D/nᵢ) > 0: E_F lies above E_i, closer to the conduction band, E_c − E_F = kT ln(N_c/N_D). p-type: acceptors N_A give p = n + N_A ≈ N_A ≫ nᵢ. Then E_i − E_F = kT ln(N_A/nᵢ) > 0: E_F lies below E_i, closer to the valence band, E_F − E_v = kT ln(N_v/N_A). The shift grows with doping and falls back toward E_i as temperature rises and nᵢ overtakes the doping.