(a) Show that the energy of the triplet state (S = 1) is not equal to the energy of the singlet state (S = 0). 10 marks (b) ρ⁰…

(a)

For a two-electron atom or excited configuration with two distinct one-electron orbitals φ₁ and φ₂, the total wavefunction must be antisymmetric under exchange of the two electrons. The singlet state S = 0 has antisymmetric spin function, so its spatial part must be symmetric: ψₛ = (1/√2)[φ₁(r₁)φ₂(r₂) + φ₁(r₂)φ₂(r₁)]. The triplet state S = 1 has symmetric spin function, so its spatial part must be antisymmetric: ψₜ = (1/√2)[φ₁(r₁)φ₂(r₂) − φ₁(r₂)φ₂(r₁)].

Take the two-electron Hamiltonian H = h₁ + h₂ + e²/(4πε₀ r₁₂). Define the direct Coulomb integral J = ∫∫ |φ₁(r₁)|² |φ₂(r₂)|² e²/(4πε₀ r₁₂) dr₁ dr₂ and the exchange integral K = ∫∫ φ₁*(r₁)φ₂*(r₂) e²/(4πε₀ r₁₂) φ₂(r₁)φ₁(r₂) dr₁ dr₂.

Using the above spatial functions: Eₛ = ⟨ψₛ|H|ψₛ⟩ = E₁ + E₂ + J + K, Eₜ = ⟨ψₜ|H|ψₜ⟩ = E₁ + E₂ + J − K. Therefore Eₛ − Eₜ = 2K. For overlapping atomic orbitals and repulsive electron-electron interaction, K ≠ 0 and is usually positive. Hence the singlet and triplet energies are not equal; the splitting is the exchange splitting. This requires the two electrons to occupy different spatial orbitals. If both electrons occupy the same spatial orbital, the antisymmetric triplet spatial function vanishes, so only the singlet state can exist.

(b)

Both ρ⁰ and K⁰ can be observed in the π⁺π⁻ channel, but their decay mechanisms are different.

ρ⁰ is a quark-antiquark resonance, e.g. (uū − dd̄)/√2, with zero strangeness. The decay ρ⁰ → π⁺ + π⁻ conserves strangeness and can proceed through the strong interaction. Strong decays have widths of order 100 MeV. For ρ⁰, Γ ≈ 150 MeV, so τ = ħ/Γ ≈ (6·58 × 10⁻²² MeV s)/(150 MeV) ≈ 4 × 10⁻²⁴ s, which is conventionally quoted as of order 10⁻²³ s, the strong-interaction timescale.

K⁰ has strangeness S = +1. The final pions have S = 0, so the decay requires ΔS = 1. Strong and electromagnetic interactions conserve strangeness, so K⁰ cannot decay strongly or electromagnetically to π⁺π⁻. It must decay through the weak interaction, which is much weaker. The 0·89 × 10⁻¹⁰ s value is essentially the lifetime of the short-lived neutral kaon component K_S. Its weak width is Γ ≈ ħ/τ ≈ (6·58 × 10⁻²² MeV s)/(0·89 × 10⁻¹⁰ s) ≈ 7·4 × 10⁻¹² MeV, which is about 10¹³ times smaller than the ρ⁰ width. Thus the huge lifetime difference is due to strong decay for ρ⁰ and strangeness-changing weak decay for K⁰.

(c)

In fcc, the body centre is an octahedral interstitial void coordinated by six face-centred atoms. Let the lattice constant be a and the atomic radius be R.

For fcc atoms touching along a face diagonal: 4R = √2 a, so R = a/(2√2).

The body centre is at (a/2, a/2, a/2). A nearest face-centred atom is, for example, at (a/2, a/2, 0). The distance from the body centre to this face-centred atom is d = a/2.

Let r be the radius of the interstitial sphere that just fits. It touches the six face-centred atoms, so d = R + r. Hence r = a/2 − R = a/2 − a/(2√2) = (a/2)(1 − 1/√2).

Using a = 2√2 R: r = √2 R − R = (√2 − 1)R. Therefore r = (√2 − 1)R ≈ 0·414R. Equivalently, r/a = (√2 − 1)/(2√2) ≈ 0·1464. This is the octahedral void radius in fcc.

(d)

Lead has an fcc structure. For a cubic crystal, the interplanar spacing for (hkl) is dₕₖₗ = a/√(h² + k² + l²). For (220), h² + k² + l² = 2² + 2² + 0² = 8, so d₂₂₀ = a/√8 = a/(2√2).

For fcc, atoms touch along the face diagonal: 4R = √2 a, so R = a/(2√2). Thus d₂₂₀ = R.

Using Bragg’s law for first order, n = 1: λ = 2 d₂₂₀ sinθ = 2R sinθ. Therefore R = λ/(2 sinθ).

Put λ = 1·54 Å and θ = 32°: sin32° = 0·5299. R = 1·54/(2 × 0·5299) = 1·54/1·0598 = 1·453 Å. R ≈ 1·45 Å. The corresponding lattice constant is a = 2√2 R ≈ 4·11 Å.

(e)(i)

Differences in electrical characteristics of JFET and MOSFET:

• Gate structure: JFET uses a reverse-biased p-n junction gate. MOSFET uses an insulated gate separated by a thin oxide layer.
• Gate current and input resistance: JFET has very small but finite gate leakage current; input resistance is high, typically about 10⁸–10¹⁰ Ω. MOSFET gate leakage is extremely small through the oxide, so input resistance is much higher, often 10¹²–10¹⁵ Ω.
• Control mechanism: JFET controls current by varying the depletion-layer width in the channel. MOSFET controls current by field-induced inversion or depletion of the channel.
• Operating mode: JFET operates only in depletion mode and is normally ON. MOSFET can be enhancement type or depletion type; the common enhancement MOSFET is normally OFF.
• Transfer characteristic: For an n-channel JFET, I_D = I_DSS(1 − V_GS/V_P)². For an enhancement n-channel MOSFET, I_D = k(V_GS − V_T)² for V_GS > V_T.
• Capacitance and noise: MOSFETs usually have higher input capacitance and greater flicker noise due to oxide traps. JFETs often have lower noise.
• Ruggedness: MOSFET gate oxide can be damaged by electrostatic discharge. JFET is relatively more robust.
• Terminals: JFET is normally a three-terminal device. MOSFET often has a fourth body/substrate terminal, giving rise to the body effect.

(e)(ii)

In an n-channel FET, current is carried by electrons; in a p-channel FET, current is carried by holes. An n-channel MOSFET conducts when V_GS is more positive than the threshold voltage, with drain positive with respect to source. A p-channel MOSFET conducts when V_GS is more negative than the threshold voltage, with drain negative with respect to source. For JFETs, an n-channel JFET requires a negative gate-to-source voltage, while a p-channel JFET requires a positive gate-to-source voltage. Since electron mobility is higher than hole mobility, an n-channel FET usually has higher transconductance, lower ON resistance, and faster switching than a comparable p-channel FET.