(a) A particle constrained to move along x-axis in the domain 0 ≤ x ≤ L has a wave function ψ(x) = sin(nπx/L), where n is an…

(a) Normalisation and ⟨pₓ⟩ of a particle in a box

Let ψ = A sin(nπx/L) with n = 1, 2, 3, …, which vanishes at x = 0 and x = L as the walls require. Normalisation: 1 = |A|² ∫₀^L sin²(nπx/L) dx = |A|² ∫₀^L (1 − cos(2nπx/L))/2 dx = |A|² L/2, so A = √(2/L) and ψₙ(x) = √(2/L) sin(nπx/L). Expectation value of momentum, with pₓ = −iħ d/dx: ⟨pₓ⟩ = ∫₀^L ψ(−iħ dψ/dx) dx = −iħ(2/L)(nπ/L) ∫₀^L sin(nπx/L) cos(nπx/L) dx = −iħ(nπ/L²) ∫₀^L sin(2nπx/L) dx = −iħ(nπ/L²) [−(L/2nπ) cos(2nπx/L)]₀^L = 0, since cos 2nπ = cos 0 = 1. So ⟨pₓ⟩ = 0. Interpretation: the stationary state is a standing wave, an equal superposition of momenta +nπħ/L and −nπħ/L, so the average cancels while ⟨p²⟩ = n²π²ħ²/L² is not zero.

(b) Most probable distance in the hydrogen 2p state

For n = 2, l = 1: R₂₁(r) = (1/(2√6)) a₀^−3/2 (r/a₀) e^−r/2a₀. The radial probability density, defined so that P(r)dr is the probability of finding the electron between r and r + dr, is P(r) = r²|R₂₁|² = r⁴ e^−r/a₀/(24a₀⁵). Maximum: dP/dr = [r³e^−r/a₀/(24a₀⁵)](4 − r/a₀) = 0, giving r = 4a₀ (r = 0 and ∞ are minima where P = 0; the second derivative is negative at 4a₀). So r_max = 4a₀ = 2.12 Å. At this distance: P(4a₀) = (4a₀)⁴e⁻⁴/(24a₀⁵) = (32/3)e⁻⁴/a₀ = 0.195/a₀ (about 3.7×10⁹ m⁻¹). For reference, the volume density is |R₂₁(4a₀)|² = (2/3)e⁻⁴/a₀³. The result does not depend on m, since P(r) integrates over the angles. The Bohr orbit radius for n = 2 is also 4a₀, but the quantum peak for 2p coincides with it only because l = n − 1.

(c) Nuclear magnetic resonance and MRI

Definition: NMR is the resonant absorption of radio-frequency radiation by nuclei with non-zero spin (¹H, ¹³C, ³¹P) placed in a static magnetic field B₀. Principle: a nucleus has magnetic moment μ = γS. In B₀ ẑ the energy is E = −γħ m B₀ with m = ±½, so the two levels differ by ΔE = γħB₀. Resonance occurs when hν = γħB₀, i.e. at the Larmor frequency ω₀ = γB₀ (for protons γ/2π = 42.58 MHz/T, so 63.9 MHz at 1.5 T). In equilibrium the slight excess in the lower level gives a net magnetisation M along B₀. An RF pulse at ω₀ tips M into the transverse plane; M then precesses and induces a signal in a receiver coil, the free-induction decay. Recovery is governed by the longitudinal relaxation time T₁ and the transverse dephasing time T₂. MRI: the patient lies in a strong B₀ (1.5 or 3 T). The hydrogen nuclei of water and fat give the signal. A linear gradient G makes the Larmor frequency position dependent, ω(r) = γ(B₀ + G·r). A slice-selective RF pulse with a narrow bandwidth excites one slice only; a phase-encoding gradient and a frequency-encoding gradient (during readout) label positions in the plane. The signal is recorded as a set of spatial frequencies (k-space), and a two-dimensional Fourier transform reconstructs the image. Contrast comes from proton density, T₁ and T₂. Short TR and short TE give T₁ weighting; long TR and long TE give T₂ weighting; long TR and short TE give proton-density weighting. Brain imaging is a standard example, where the grey/white-matter contrast and tumours are shown without ionising radiation.