University Quantum Mechanics II · Quantum Scattering Theory · 15.1
Cross-sections and the classical benchmarks
Before any wavefunction appears, this lesson settles what a scattering experiment actually measures and hands you the two classical answers the rest of the unit is checked against. Learn to turn a count into an area, and to read an orbit's b(θ) as a cross-section.
Build the model
Connect the measurement to the mechanism.
A scattering experiment measures nothing but counts, and the cross-section is how a count becomes physics: divide the rate into a detector by the incident flux and by the detector's solid angle, and what remains is an area per steradian that belongs to the interaction alone. Classical mechanics can compute it without a wave in sight. Each particle's impact parameter b fixes its deflection θ, and conserving particles from the annulus 2πb db into the cone 2π sin θ dθ gives dσ/dΩ = (b/sin θ)|db/dθ|, whose integral is the area of beam deflected at all.
Two orbits are solvable exactly, and the rest of the unit is measured against them. The hard sphere gives R²/4 at every angle and a total πR², its own shadow. The Coulomb hyperbola gives Rutherford's (d/4)²/sin⁴(θ/2), of order α² with ħ cancelling, and a total that diverges because a 1/r force bends every orbit however far away it passes.
Both are worked in the centre-of-mass frame with the reduced mass m. They hold while the wavelength is short against the scale of the force: η = kd/2 ≫ 1 for Coulomb orbits, and kR ≫ 1 for the sphere, outside a forward diffraction peak about 1/kR wide. Past that, the benchmarks part company: the quantum hard sphere scatters four times its classical total as kR → 0, while the Rutherford law survives quantisation exactly, a coincidence special to 1/r (both quoted here and met later in the unit).
- Simple definition
- The differential cross-section dσ/dΩ is the rate at which particles scatter into unit solid angle about a direction, divided by the incident flux; it is an area per steradian that belongs to the interaction, and its integral σ is the beam area the target removes.
- Example
- 10⁶ alphas per second cross a gold foil holding 5.9 × 10²² atoms m⁻². A 1.0 msr detector at 90° counting 110 an hour, 0.0306 s⁻¹, measures dσ/dΩ = 0.0306/(10⁶ × 5.9 × 10²² × 10⁻³) = 5.2 × 10⁻²⁸ m² sr⁻¹, or 5.2 b sr⁻¹.
Defined here. Dividing out the flux and the detector leaves a property of the interaction alone; σ = ∫(dσ/dΩ) dΩ is the beam area the target removes.
dN/dt in s⁻¹ into solid angle dΩ in sr; J incident flux in m⁻² s⁻¹; dσ/dΩ in m² sr⁻¹, or b sr⁻¹ with 1 b = 10⁻²⁸ m²
Derived here by summing the counting definition over the n t A atoms the beam crosses. Valid while n t σ ≪ 1, so each particle scatters at most once.
I beam rate in s⁻¹; n target atoms in m⁻³; t foil thickness in m; ΔΩ detector solid angle in sr
Derived here: the beam annulus 2πb|db| feeds the cone 2π sin θ |dθ|. Integrated, it returns σ = ∫2πb db, finite only if the force has a finite range.
b impact parameter in m; θ CM deflection in rad; add the terms when several b reach the same θ
Derived here from mirror reflection. Isotropic in the CM frame, with total σ = πR², exactly the geometric shadow; no α and no ħ appear.
R is the contact distance between centres in m, r₁ + r₂ for two spheres; dσ/dΩ in m² sr⁻¹
Derived here from the impulse along the orbit's axis. Order α², since d = Z₁Z₂αħc/E, with ħ cancelling; the exact quantum result is identical (quoted).
d head-on closest approach in m; E CM kinetic energy in J; e²/(4πε₀) = 1.440 MeV fm; b = (d/2) cot(θ/2)
Derived here: every b below b(θ₀) is deflected by at least θ₀. It diverges as θ₀ → 0 because a 1/r force bends every orbit; atomic screening makes it finite.
θ₀ the smallest deflection counted; σ in m²; for small θ₀ it grows as πd²/θ₀²
Counting turns a rate into an area
A scattering experiment only counts. A beam of I particles per second spread evenly over an area A has flux J = I/A, and a detector at angle θ subtending dΩ records dN/dt. For one scatterer that rate must be proportional to J and to dΩ, so dσ/dΩ = (dN/dt)/(J dΩ) depends on the interaction alone, with dimensions s⁻¹ ÷ (m⁻² s⁻¹ sr) = m² sr⁻¹. That is a definition, not a result. A real target is a foil with n atoms per m³ and thickness t, so the beam meets n t A scatterers and A cancels: dN/dt = I n t (dσ/dΩ) ΔΩ. The sum assumes a thin foil, n t σ ≪ 1, so that each particle scatters at most once, and atoms that scatter independently. The total σ = ∫(dσ/dΩ) dΩ is the removal rate divided by the flux, the area of beam the target takes out; applied slice by slice through a thick target it gives I(x) = I₀ e(−nσx). Nuclear cross-sections come in barns, 1 b = 10⁻²⁸ m² = 100 fm². No detector can sit in the unscattered beam, so θ = 0 is never counted directly.
From an orbit to a cross-section
Classically each particle follows a definite orbit fixed by its impact parameter b, the perpendicular distance of its incoming line from the force centre, and a central force keeps that orbit in one plane through the beam axis. The particles crossing the annulus between b and b + db, within azimuth dφ, therefore leave between θ and θ + dθ in the same plane. Counting them both ways, J b|db| dφ = J (dσ/dΩ) sin θ |dθ| dφ, so dσ/dΩ = (b/sin θ)|db/dθ|, derived here. The modulus is there because db/dθ is negative for a repulsive force: the larger b, the gentler the deflection. If several impact parameters reach one θ, their terms add. Two features make the formula infinite: a rainbow angle, where dθ/db = 0 and a band of b crowds into one direction, and a glory, where sin θ = 0 at nonzero b; waves smooth both. Integrate over angle and the formula returns σ = ∫2πb db, the area of beam deflected at all. That is finite only if some bₘₐₓ exists beyond which θ is exactly zero, which is to say only if the force has a finite range.
The hard sphere: isotropic, and exactly its shadow
An impenetrable sphere of radius R reflects a point particle like a mirror. A particle with b ≤ R strikes where the radius makes an angle χ with the beam, b = R sin χ, and leaves at θ = π − 2χ, so b = R cos(θ/2) and |db/dθ| = (R/2) sin(θ/2). Then dσ/dΩ = R cos(θ/2) × (R/2) sin(θ/2)/sin θ = R²/4, derived here, the same at every angle, and σ = 4π × R²/4 = πR², the shadow, since every b ≤ R scatters and no larger b does. Two spheres touch when their centres are r₁ + r₂ apart, so that sum is R, and identical molecules of diameter D give σ = πD². No α and no ħ enter: the model is geometry. Measured check: gas-viscosity tables give σ = 0.43 nm² for N₂, so D = 0.37 nm, against 0.39 nm from the volume per molecule of liquid nitrogen. The model needs short waves. N₂ pairs at 300 K have kR ≈ 60, but helium at 4 K has kR ≈ 2, where the wave treatment of topic 15.06 takes over, reaching 4πR² as kR → 0.
Rutherford's orbit from one impulse
For V = Z₁Z₂e²/(4πε₀r), define d = Z₁Z₂e²/(4πε₀E), the head-on distance of closest approach, with E = ½mv² the CM kinetic energy. The momentum turns through θ at fixed size mv, so it changes by 2mv sin(θ/2) along the orbit's symmetry axis. Summing the force's component along that axis, with dt = mr² dψ/L for the orbit's polar angle ψ and L = mvb, gives [Z₁Z₂e²/(4πε₀)](m/L) × 2cos(θ/2). Equating the two: tan(θ/2) = d/(2b), so b = (d/2) cot(θ/2), the hyperbola's own result. Then |db/dθ| = (d/4)/sin²(θ/2), and the orbit formula gives dσ/dΩ = (d/4)²/sin⁴(θ/2), derived here. Checks: (d/4)² is an area; at θ = π it is (d/4)², and at small θ it becomes d²/θ⁴. Writing d = Z₁Z₂αħc/E shows the result is of order α², with ħ cancelling because αħc = e²/(4πε₀) = 1.440 MeV fm. For alphas on gold at 5.00 MeV, d = 45.5 fm, and dσ/dΩ falls from 2.24 × 10⁴ b sr⁻¹ at 10° to 1.29 b sr⁻¹ at 180°. Geiger and Marsden counted this 1/sin⁴(θ/2) law from 5° to 150° in 1913; Chadwick's 1920 absolute counts gave nuclear charges 29.3, 46.3 and 77.4 for Cu, Ag and Pt, against Z = 29, 47 and 78 (measured, quoted).
An infinite total, and what the atom does to it
Every b below b(θ₀) is deflected by at least θ₀, so σ(θ ≥ θ₀) = πb(θ₀)² = (πd²/4) cot²(θ₀/2), derived here. For alphas on gold at 5.00 MeV that is 16.3 b beyond 90°, 2.1 × 10³ b beyond 10° and 2.1 × 10⁵ b beyond 1°, growing as 1/θ₀² with no limit. The cause is range, not strength: however large b is, a 1/r potential still bends the orbit by θ ≈ d/b, so no bₘₐₓ exists. Real nuclei sit inside atoms whose electrons screen the charge beyond roughly the Thomas–Fermi radius a = 0.885a₀Z(−1/3), 10.9 pm for gold (quoted from atomic physics). Cutting b off there leaves no deflection smaller than about θ ≈ d/a = 4.2 mrad and caps the total near πa² ≈ 4 × 10⁶ b, a crude but finite figure set by the atom rather than the nucleus. The orbit picture itself needs the Sommerfeld parameter η = kd/2 ≫ 1, and these alphas have η = 22. Yet the Rutherford law does not depend on it: the exact quantum Coulomb problem, quoted here and treated at Born level in topic 15.09, returns the same |f|² at every η.
Centre-of-mass angles and laboratory angles
Everything above is in the CM frame. Unit 2's reduction replaces the two bodies by one particle of reduced mass m = m₁m₂/(m₁ + m₂) moving in V(r) at the relative speed v, so E = ½mv² is the CM kinetic energy, a fraction m₂/(m₁ + m₂) of the laboratory energy for a target at rest, and θ is the angle through which the relative velocity turns. A laboratory detector measures the projectile's angle θL. For elastic scattering, adding back the CM velocity gives tan θL = sin θ/(cos θ + γ) with γ = m₁/m₂, and the counts transform as dσ/dΩL = (dσ/dΩ)(1 + 2γ cos θ + γ²)³⁄²/|1 + γ cos θ|, both quoted from classical kinematics. As γ → 0 both reduce to the CM forms, and totals never change: a particle scattered in one frame is scattered in every frame. For alphas on gold γ = 0.0203, so ECM = 0.980 Elab, 90° in the CM is 88.8° in the lab, and the solid-angle factor there is 1.0006. For equal masses θL = θ/2 and nothing goes backward in the lab, which is where the conversion stops being a detail.
Change one variable at a time
Make the relationship visible.
Walk θ₀ from 175° down to 5°: the sphere's σ(θ ≥ θ₀) creeps up to πR² and stops, while Rutherford's grows as 1/θ₀² without limit. Then double E: the solid curve drops 0.6 of a decade at every angle and keeps its shape, because dσ/dΩ ∝ 1/E².
RUTHERFORD b(θ₀)39.40 fm
RUTHERFORD dσ/dΩ(θ₀)20.70 b sr⁻¹
RUTHERFORD σ(θ ≥ θ₀)48.78 b
HARD SPHERE σ(θ ≥ θ₀)2.356 b
Live interpretationRUTHERFORD b(θ₀): 39.40 fm. RUTHERFORD dσ/dΩ(θ₀): 20.70 b sr⁻¹. RUTHERFORD σ(θ ≥ θ₀): 48.78 b. HARD SPHERE σ(θ ≥ θ₀): 2.356 b
Catch the common trap
Explain before calculating.
A beam of point particles strikes a fixed, impenetrable sphere of radius R and rebounds elastically. Two identical small detectors at the same distance from the sphere sit at 20° and at 160° from the beam direction. How do their count rates compare?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAlpha particles with centre-of-mass energy 6.00 MeV scatter from silver nuclei (Z = 47). Find the head-on distance d, the impact parameter that gives a 45° deflection, and dσ/dΩ at 45° in barns per steradian, once from (b/sin θ)|db/dθ| and once from the Rutherford formula. Show where the fine-structure constant enters, and check that the alpha stays clear of the nucleus.
- d = Z₁Z₂e²/(4πε₀E) = 2 × 47 × 1.440 MeV fm ÷ 6.00 MeV = 22.56 fm. Written as Z₁Z₂αħc/E it is 94 × 197.33 MeV fm ÷ (137.04 × 6.00 MeV) = 22.56 fm: one power of α.
- b = (d/2) cot(θ/2) = 11.28 fm × cot 22.5° = 11.28 fm × 2.414 = 27.23 fm.
- Orbit route: |db/dθ| = (d/4)/sin²(22.5°) = 5.640 fm ÷ 0.14645 = 38.51 fm per rad, and b/sin θ = 27.23 fm ÷ 0.7071 = 38.51 fm, so dσ/dΩ = 38.51 × 38.51 = 1483 fm² sr⁻¹.
- Rutherford route: (d/4)²/sin⁴(22.5°) = 31.81 fm² ÷ 0.021447 = 1483 fm² sr⁻¹. With 1 b = 100 fm², that is 14.83 b sr⁻¹. The two routes agree, as they must, since the formula came from the orbit.
- Checks: the result is an area per steradian; it goes as d², so as α², with ħ cancelling. Energy and angular momentum conservation give r² − dr − b² = 0 at closest approach, so rₘᵢₙ = (d/2)(1 + 1/sin(θ/2)) = 40.8 fm, far outside a silver nucleus of radius about 5.7 fm (1.2A¹⁄³ fm, quoted): the point-charge force holds all along the orbit.
Answerd = 22.6 fm, b = 27.2 fm, and dσ/dΩ = 1.48 × 10³ fm² sr⁻¹ = 14.8 b sr⁻¹ by both routes; the result is of order α² and contains no ħ.
MediumA beam of 2.0 × 10⁵ alpha particles per second, at CM energy 5.00 MeV, crosses a gold foil 0.50 μm thick with 5.91 × 10²⁸ atoms m⁻³. Find the count rate in a 1.5 msr detector at 120°, the fraction of alphas deflected beyond 90°, and the single-scattering prediction for the fraction deflected beyond 1°. Say what the last number tells you.
- Areal density n t = 5.91 × 10²⁸ m⁻³ × 0.50 × 10⁻⁶ m = 2.955 × 10²² m⁻². Head-on distance d = 2 × 79 × 1.440 MeV fm ÷ 5.00 MeV = 45.50 fm.
- dσ/dΩ(120°) = (d/4)²/sin⁴60°, with (d/4)² = (11.376 fm)² = 129.41 fm² and sin⁴60° = 0.5625, so dσ/dΩ = 230.1 fm² sr⁻¹ = 2.30 × 10⁻²⁸ m² sr⁻¹.
- Rate = I n t (dσ/dΩ) ΔΩ = 2.0 × 10⁵ s⁻¹ × 2.955 × 10²² m⁻² × 2.30 × 10⁻²⁸ m² sr⁻¹ × 1.5 × 10⁻³ sr = 2.04 × 10⁻³ s⁻¹, about 7.3 counts an hour. Units: s⁻¹ × m⁻² × m² sr⁻¹ × sr = s⁻¹.
- Beyond 90°: every b below b(90°) = (d/2) cot 45° = 22.75 fm qualifies, so σ = π(22.75 fm)² = 1626 fm² = 1.63 × 10⁻²⁷ m², and the fraction is n t σ = 4.80 × 10⁻⁵, about 1 alpha in 21 000.
- Beyond 1°: b(1°) = 22.75 fm × cot 0.5° = 2607 fm, so σ = 2.14 × 10⁻²³ m² and n t σ = 0.63. A single-scattering fraction that large breaks the thin-target condition n t σ ≪ 1: most alphas take several small kicks, and the formula cannot be trusted there. Setting n t σ = 0.1 gives cot²(θ/2) = 2081, so θ = 2.5° is roughly the smallest angle it holds for in this foil.
- Screening is not the culprit at 1°: b = 2.6 pm is a quarter of the Thomas–Fermi radius 0.885a₀Z(−1/3) = 10.9 pm (quoted), so the nuclear charge is only partly screened there, and screening weakens the deflection, so it would lower n t σ, not raise it.
AnswerAbout 2.0 × 10⁻³ s⁻¹, or 7 counts an hour, at 120°; 4.8 × 10⁻⁵ of the alphas are deflected beyond 90°; the single-scattering estimate beyond 1° is 0.63, too large to be a probability of one event, so multiple scattering dominates below about 2.5°.
HardGas-viscosity tables give the N₂–N₂ collision cross-section as 0.43 nm² (measured, quoted). Model the molecules as hard spheres, fire tagged projectile molecules at molecules at rest, count only the projectiles, and treat the two masses as equal. Find the contact distance R and the CM dσ/dΩ; derive the laboratory angle and dσ/dΩL; find the fraction of scattered projectiles within 30° of the beam; and test the classical model with kR at 300 K.
- σ = πR² gives R = √(0.43 nm²/π) = 0.370 nm, the molecular diameter. In the CM frame dσ/dΩ = R²/4 = 3.42 × 10⁻²⁰ m² sr⁻¹ at every angle. Check against a second measurement: liquid nitrogen at 0.807 g cm⁻³ has 5.76 × 10⁻²⁹ m³ per molecule, whose cube root is 0.386 nm, within 5%.
- Equal masses: the CM moves at v/2, and in the CM the projectile keeps speed v/2 and turns through θ. Adding the CM velocity back, its lab velocity is (v/2)(1 + cos θ, sin θ), so tan θL = sin θ/(1 + cos θ) = tan(θ/2) and θL = θ/2. No projectile is scattered backward in the lab.
- Solid angles: dΩ = 2π sin θ dθ = 2π (2 sin θL cos θL)(2 dθL) = 4 cos θL dΩL. The same particles fill both, so dσ/dΩL = (R²/4) × 4 cos θL = R² cos θL = 1.37 × 10⁻¹⁹ cos θL m² sr⁻¹. The quoted general factor (1 + 2γ cos θ + γ²)³⁄²/|1 + γ cos θ| at γ = 1 is 4 cos(θ/2), the same.
- Total check: ∫ R² cos θL × 2π sin θL dθL from 0 to 90° = πR² = 0.43 nm², the CM total, as it must be. Within 30°: 2πR² × ½ sin²30° over πR² gives sin²30° = 0.25; in the CM that is θ ≤ 60°, holding (1 − cos 60°)/2 = 0.25 of an isotropic distribution. Both frames agree.
- Validity: the reduced mass is 14.0 u = 2.33 × 10⁻²⁶ kg; at E = (3/2)kB T = 6.21 × 10⁻²¹ J, k = √(2mE)/ħ = 1.61 × 10¹¹ m⁻¹, so kR ≈ 60 and the wavelength, 0.039 nm, is a tenth of R: classical orbits are fair. Helium at 4.2 K, with σ = 0.21 nm², gives kR ≈ 1.9, and the classical model fails.
AnswerR = 0.370 nm and CM dσ/dΩ = 3.42 × 10⁻²⁰ m² sr⁻¹; θL = θ/2 and dσ/dΩL = R² cos θL = 1.37 × 10⁻¹⁹ cos θL m² sr⁻¹; 25% of scattered projectiles lie within 30° of the beam; kR ≈ 60 at 300 K, so the classical picture holds.