---
title: "Scattering Theory"
book: "University Physics — Year 3"
subject: physics
language: en
chapter: 15
exercises: 12
source: https://one-course.com/books/physics/5/en/chapter/15-scattering-theory
---

# Chapter 15 — Scattering Theory

Almost everything we know about the small is learned by throwing things at it and watching what bounces. Rutherford discovered the nucleus by counting alpha-particle flashes on a zinc-sulfide screen; fifty years later the same experiment, run with electrons at higher energy, found quarks inside the proton; between and since, every [cross-section](#def-b3-scattering-theory-cross-section) in nuclear, particle, atomic and condensed-matter physics has been read through the theory of this chapter. The language is the *[cross-section](#def-b3-scattering-theory-cross-section)* — an effective target area; the quantum object is the *[scattering amplitude](#thm-b3-scattering-theory-amplitude)*, whose [Born approximation](#thm-b3-scattering-theory-born) says something unforgettable: a diffuse target is probed in Fourier space, angle by angle. At low energy the [partial waves](#thm-b3-scattering-theory-partial) take over, collapsing every complicated potential into a single number — the [scattering length](#thm-b3-scattering-theory-partial) — that today tunes ultracold quantum gases; and where a projectile can linger in a quasi-bound state, the [cross-section](#def-b3-scattering-theory-cross-section) spikes into resonances, the spectroscopy of the otherwise invisible.

## 15.1 Cross-sections

**Definition 15.1 (Cross-section).**

Send a uniform beam of flux $\Phi$ (particles per unit area per second) onto one target particle. The rate of projectiles scattered into the solid angle $\dd\Omega$ around the direction $(\theta,
\varphi)$ defines the *differential cross-section*,

$$
\dd\dot N = \Phi\,\frac{\dd\sigma}{\dd\Omega}\,\dd\Omega ,
$$

and its integral $\sigma = \int(\dd\sigma/\dd\Omega)\,\dd\Omega$ is the *total cross-section*: the area the target effectively presents. For a thin slab with $n$ targets per unit volume, a beam attenuates as $\eu^{-n\sigma x}$: cross-sections are measured by counting what emerges. Units: nuclear physics uses the *barn*, $1\,\mathrm{b} = 10^{-28}\,\mathrm{m}^{2}$ — about a uranium nucleus’s geometric size, “as big as a barn” to a neutron.

![The scattering experiment: a plane wave in, a spherical wave out, a detector at angle counting. The angular pattern of the counts is the physics of the target.](https://one-course.com/images/onecourse/chapters/physics-5/b3-scattering-theory/fig-c02e790bd1cf.svg)

*The scattering experiment: a plane wave in, a spherical wave out, a detector at angle $\theta$ counting. The angular pattern of the counts is the physics of the target.*

## 15.2 The amplitude and the Born approximation

**Theorem 15.2 (Scattering states and the amplitude).**

For a particle of energy $E = \hbar^2k^2/2m$ meeting a localised potential $V(\vect r)$, the stationary scattering state behaves at large distance as

$$
\psi(\vect r) \;\longrightarrow\;
\eu^{\iu kz} + f(\theta)\,\frac{\eu^{\iu kr}}{r} ,
$$

an incident plane wave plus an outgoing spherical wave modulated by the *[scattering amplitude](#thm-b3-scattering-theory-amplitude)* $f(\theta)$ (a length), and

$$
\frac{\dd\sigma}{\dd\Omega} = |f(\theta)|^2 .
$$

**Partial proof.** The $1/r$ fall-off makes the outgoing flux through any large sphere finite; computing the [probability currents](https://one-course.com/books/physics/5/en/chapter/7-the-schrodinger-equation-in-three-dimensions#prop-b3-schrodinger-three-dimensions-current) ([Proposition 7.2](https://one-course.com/books/physics/5/en/chapter/7-the-schrodinger-equation-in-three-dimensions#prop-b3-schrodinger-three-dimensions-current)) of the two terms, the radial outgoing current per solid angle is $(\hbar k/m)|f|^2$ against the incident flux $\hbar k/m$: the ratio is the definition of $\dd\sigma/\dd\Omega$. That solutions of this asymptotic form exist is admitted. ∎

**Theorem 15.3 (Born approximation).**

For a weak potential (or fast projectile), first-order [perturbation theory](https://one-course.com/books/physics/5/en/chapter/13-perturbation-theory#thm-b3-perturbation-theory-corrections) in $V$ gives

$$
f(\theta) = -\frac{m}{2\pi\hbar^2}
\int V(\vect r)\,\eu^{\iu\vect q\cdot\vect r}\,\dd^3r ,
\qquad
\vect q = \vect k - \vect k' , \quad
q = 2k\sin\frac\theta2 :
$$

the amplitude is, up to constants, the *Fourier transform of the potential* evaluated at the [momentum transfer](#thm-b3-scattering-theory-born) $\hbar\vect q$. Scattering experiments are Fourier analysers of matter: small angles read long wavelengths (coarse structure), large angles the fine detail — the deepest reason why “higher energy” means “smaller distances”.

**Partial proof.** This is Fermi’s golden rule ([Theorem 13.5](https://one-course.com/books/physics/5/en/chapter/13-perturbation-theory#thm-b3-perturbation-theory-golden)) run between plane waves $\ket{\vect k} \to \ket{\vect k'}$: the matrix element is $\int V\eu^{\iu(\vect k - \vect k')\cdot\vect r}\dd^3r$, the density of final states supplies the kinematic factors, and the bookkeeping (admitted) assembles them into $f$. Elasticity gives $|\vect k'| = |\vect k|$, whence $q = 2k\sin(\theta/2)$ by the isoceles triangle of $\vect k, \vect k'$. ∎

**Example 15.4 (From Yukawa to Rutherford).**

For the screened Coulomb potential $V = \dfrac{Q_1Q_2}{4\pi
\varepsilon_0 r}\,\eu^{-r/a}$ the Born integral is elementary ([Exercise 15.5](#exo-b3-scattering-theory-5)) and, as the screening radius $a \to \infty$,

$$
\frac{\dd\sigma}{\dd\Omega} =
\Big(\frac{Q_1Q_2}{16\pi\varepsilon_0 E}\Big)^2
\frac{1}{\sin^4(\theta/2)} :
$$

the *Rutherford [cross-section](#def-b3-scattering-theory-cross-section)*. By a coincidence unique to $1/r$, the classical calculation, the [Born approximation](#thm-b3-scattering-theory-born) and the exact quantum answer all agree — which is why Rutherford, computing classically in 1911, got the right formula and, from its verification flash by flash, the nucleus ([Problem 15.1](#pb-b3-scattering-theory-1)).

## 15.3 Partial waves and the scattering length

**Theorem 15.5 (Phase shifts).**

For a central potential, decompose the scattering state over angular momenta: each partial wave $l$ leaves the potential region with its radial oscillation *shifted* by a phase $\delta_l(k)$ — attraction advances it, repulsion delays it — and the observable amplitude reassembles as

$$
f(\theta) = \frac1k\sum_{l=0}^\infty(2l+1)\,\eu^{\iu\delta_l}
\sin\delta_l\,P_l(\cos\theta) , \qquad
\sigma = \frac{4\pi}{k^2}\sum_l(2l+1)\sin^2\delta_l .
$$

At low energy, classical intuition ($l \sim kb$ for impact parameter $b$) says only $l = 0$ penetrates: scattering becomes isotropic and one number rules,

$$
f \to -a , \qquad \sigma \to 4\pi a^2 ,
$$

where $a = -\lim_{k\to0}\delta_0/k$ is the *[scattering length](#thm-b3-scattering-theory-partial)*. Whatever its inner complexity, a target at low energy *is* its [scattering length](#thm-b3-scattering-theory-partial) — the great simplification on which cold-atom physics runs.

**Proof.** *Admitted at this level.* ∎

![A partial wave leaves the interaction region with its oscillation displaced: the phase shift _0. Everything a detector can know about a central potential at one energy is the list of these shifts.](https://one-course.com/images/onecourse/chapters/physics-5/b3-scattering-theory/fig-e2fba7fd2b97.svg)

*A partial wave leaves the interaction region with its oscillation displaced: the [phase shift](#thm-b3-scattering-theory-partial) $\delta_0$. Everything a detector can know about a central potential at one energy is the list of these shifts.*

**Example 15.6 (Neutron meets proton).**

Slow neutrons scatter off protons with $\sigma \approx
20.4\,\mathrm{b}$ — an area forty times the deuteron’s geometric size. The explanation: the low-energy collision is pure $s$ wave with two spin channels, whose measured [scattering lengths](#thm-b3-scattering-theory-partial), $a_{\text{t}} = 5.4\,\mathrm{fm}$ (triplet) and $a_{\text{s}} =
-23.7\,\mathrm{fm}$ (singlet), are both far larger than the $2\,\mathrm{fm}$ force range. Large $|a|$ means a state *almost* at zero energy: the triplet’s barely bound deuteron ([Example 7.9](https://one-course.com/books/physics/5/en/chapter/7-the-schrodinger-equation-in-three-dimensions#ex-b3-schrodinger-three-dimensions-deuteron)), and in the singlet a “virtual” partner that just fails to bind — its failure written in $a_{\text{s}} < 0$. A twenty-barn puzzle, solved by two nearly-there bound states ([Exercise 15.7](#exo-b3-scattering-theory-7)).

## 15.4 Resonances

**Proposition 15.7 (Breit–Wigner resonances).**

If the projectile can be temporarily captured into a quasi-bound state of energy $E_{\text{R}}$ and lifetime $\tau = \hbar/\Gamma$, the corresponding [phase shift](#thm-b3-scattering-theory-partial) sweeps through $\pi/2$ and the partial [cross-section](#def-b3-scattering-theory-cross-section) spikes:

$$
\sigma_l(E) \approx \frac{4\pi}{k^2}\,(2l+1)\,
\frac{\Gamma^2/4}{(E - E_{\text{R}})^2 + \Gamma^2/4} ,
$$

a Lorentzian peak of width $\Gamma$ at the ceiling the wave can reach (the “unitarity limit” $4\pi(2l+1)/k^2$). Read backwards, a resonance in a [cross-section](#def-b3-scattering-theory-cross-section) *is* the discovery of a state: its position is an energy level, its width a lifetime. Most of the particle zoo of [Chapter 26](https://one-course.com/books/physics/5/en/chapter/26-particle-physics#ch-b3-particle-physics) was discovered as exactly such bumps.

**Proof.** *Admitted at this level.* ∎

![Two faces of the phase shift. Left: a quasi-bound state sweeps through π/2 — a Breit–Wigner peak whose width is a lifetime. Right: the Ramsauer–Townsend effect — in argon, _0 passes through π near 1\, eV and the atom turns almost transparent to electrons: destructive interference as an anti-resonance.](https://one-course.com/images/onecourse/chapters/physics-5/b3-scattering-theory/fig-c4132b1b159b.svg)

*Two faces of the [phase shift](#thm-b3-scattering-theory-partial). Left: a quasi-bound state sweeps $\delta$ through $\pi/2$ — a Breit–Wigner peak whose width is a lifetime. Right: the Ramsauer–Townsend effect — in argon, $\delta_0$ passes through $\pi$ near $1\,\mathrm{eV}$ and the atom turns almost transparent to electrons: destructive interference as an anti-resonance.*

**Method 15.8 (Reading a scattering problem).**

(1) Kinematics first: $k$, and the accessible $q$-range $0 \le q
\le 2k$ — what structure scales are visible? (2) Fast and weak: Born — Fourier-transform the potential; check the smallness. (3) Slow: [partial waves](#thm-b3-scattering-theory-partial), usually $s$ only; think [scattering length](#thm-b3-scattering-theory-partial), and expect $\sigma = 4\pi a^2$ (bosonic [identical particles](https://one-course.com/books/physics/5/en/chapter/14-identical-particles#def-b3-identical-particles-exchange): $8\pi a^2$). (4) Peaks in $\sigma(E)$: fit Breit–Wigner, report a level and a lifetime; dips: a phase passing $\pi$. (5) Extended targets: multiply the point amplitude by the form factor — the Fourier transform of the density ([Exercise 15.12](#exo-b3-scattering-theory-12)).

![The first scattering laboratory: a brass microscope aimed at a zinc-sulfide screen, counting one faint green flash per arriving particle. From such counts, angle by angle, Rutherford weighed the atom’s core.](https://one-course.com/images/onecourse/chapters/physics-5/b3-scattering-theory/img-340462bc646f.jpg)

*The first scattering laboratory: a brass microscope aimed at a zinc-sulfide screen, counting one faint green flash per arriving $\alpha$ particle. From such counts, angle by angle, Rutherford weighed the atom’s core.*

## 15.5 Exercises

**Exercise 15.1 ★.**

A $1\,\text{µ}\mathrm{A}$ proton beam ($6.2 \times 10^{12}$ protons/s) of $1\,\mathrm{cm}^{2}$ section crosses a $1\,\text{µ}\mathrm{m}$ gold foil ($n =
5.9 \times 10^{28}\,\mathrm{m}^{-3}$). (a) The areal target density. (b) With $\sigma = 10\,\mathrm{b}$ per nucleus, what fraction of the beam scatters? (c) The rate into a detector of solid angle $10^{-3}\,\mathrm{sr}$ if scattering were isotropic. (d) Why is “[cross-section](#def-b3-scattering-theory-cross-section)” a well-defined property of the *pair* (projectile, target, energy) rather than of the target alone?

**Solution of Exercise 15.1.**

(a) $nt = 5.9 \times 10^{22}\,\mathrm{m}^{-2}$. (b) $nt\sigma = 5.9 \times 10^{22}
\times 10^{-27} = 5.9 \times 10^{-5}$. (c) Scattered rate $6.2 \times 10^{12} \times 5.9 \times 10^{-5} = 3.7 \times 10^{8}\,\mathrm{s}^{-1}$; into $10^{-3}/4\pi$ of the sphere: $2.9 \times 10^{4}\,\mathrm{s}^{-1}$. (d) The “area” encodes the interaction and the wavelength: the same gold nucleus presents different areas to alphas, electrons and neutrons, and at different energies.

**Exercise 15.2 ★.**

Classical hard spheres of radii $R_1, R_2$: (a) show $\sigma =
\pi(R_1 + R_2)^2$. (b) For air molecules ($R \approx
0.15\,\mathrm{nm}$, $n = 2.5 \times 10^{25}\,\mathrm{m}^{-3}$), the mean free path $1/n\sigma$: evaluate, and compare with the kinetic-theory value of the Year 1 volume. (c) Why does a classical hard sphere show no angular structure ($\dd\sigma/\dd\Omega$ constant — accept or derive)? (d) Which feature of *quantum* hard-sphere scattering ([Exercise 15.6](#exo-b3-scattering-theory-6)) has no classical counterpart?

**Solution of Exercise 15.2.**

(a) Centres closer than $R_1 + R_2$ collide: $\sigma = \pi(R_1 +
R_2)^2$. (b) $\sigma \approx 2.8 \times 10^{-19}\,\mathrm{m}^{2}$: $\ell = 1/n\sigma
\approx 0.14\,\text{µ}\mathrm{m}$ — the kinetic-theory scale (the $\sqrt2$ of relative motion refines it). (c) A hard sphere reflects each impact-parameter ring uniformly over angle: isotropic in the classical limit. (d) The factor-of-four [cross-section](#def-b3-scattering-theory-cross-section) at low energy — pure wave diffraction, no classical shadow logic survives it.

**Exercise 15.3 ★.**

[Momentum transfer](#thm-b3-scattering-theory-born). (a) Draw the triangle of $\vect k$, $\vect k'$ and derive $q = 2k\sin(\theta/2)$. (b) For $5.5\,\mathrm{MeV}$ alphas ($k = 1.0 \times 10^{15}\,\mathrm{m}^{-1}$): the range of $q$ over $\theta \in
[0, \pi]$, and the smallest resolvable structure $\sim 1/q_{\max}$. (c) For $500\,\mathrm{MeV}$ electrons ($k \approx E/\hbar c$): the same. (d) Read off the moral connecting beam energy to microscope power.

**Solution of Exercise 15.3.**

(a) Isoceles triangle with apex angle $\theta$: $q =
2k\sin(\theta/2)$. (b) $q$ runs from $0$ to $2k = 2 \times 10^{15}\,\mathrm{m}^{-1}$: structures down to $\sim5 \times 10^{-16}\,\mathrm{m}$ are in principle encoded (the Coulomb barrier, in practice, keeps alphas outside). (c) $k = E/\hbar c = 2.5 \times 10^{15}\,\mathrm{m}^{-1}$: resolution $\sim2 \times 10^{-16}\,\mathrm{m}$ — inside the proton. (d) Resolution $\sim 1/q_{\max} \sim \hbar/p$: buying energy is buying a shorter ruler.

**Exercise 15.4 ★.**

From [Theorem 15.2](#thm-b3-scattering-theory-amplitude): (a) check that $|f|^2$ has the dimensions of an area per solid angle; (b) show the scattered flux through a sphere is $\Phi\sigma$; (c) why must $f$ generally be complex (which conservation law does its phase guard)? (d) State the optical theorem $\sigma_{\text{tot}} =
(4\pi/k)\operatorname{Im}f(0)$ and its meaning (the forward wave must be depleted by exactly what scatters away).

**Solution of Exercise 15.4.**

(a) $f$ is a length; $|f|^2$ an area — per steradian by construction. (b) Integrate $(\hbar k/m)|f|^2/r^2$ over the sphere: $\Phi\sigma$. (c) Probability conservation: the outgoing wave must interfere *destructively* with the forward beam to pay for what scatters; that bookkeeping lives in $f$’s phase. (d) The optical theorem states exactly that shadow-audit: total removal $=$ forward interference deficit.

**Exercise 15.5 ★★.**

Born for Yukawa. With $V(r) = \dfrac{g\,\eu^{-\mu r}}{r}$: (a) carry out the Born integral (angular part first, then $\int_0^
\infty\sin(qr)\eu^{-\mu r}\dd r$) to get $f = -\dfrac{2mg}
{\hbar^2(q^2 + \mu^2)}$. (b) Let $\mu \to 0$ with $g =
Q_1Q_2/4\pi\varepsilon_0$ and recover Rutherford. (c) Why does the *total* Rutherford [cross-section](#def-b3-scattering-theory-cross-section) diverge, and which physical ingredient (screening by atomic electrons) restores a finite answer? (d) In particle physics the Yukawa form with $\mu =
m_\pi c/\hbar$ models the nuclear force: compute its range $1/\mu$ for $m_\pi c^2 = 140\,\mathrm{MeV}$.

**Solution of Exercise 15.5.**

(a) The angular integral gives $4\pi\sin(qr)/qr$; then, since $\int_0^\infty\eu^{-\mu r}\sin(qr)\,\dd r = q/(q^2 + \mu^2)$, one finds $f = -2mg/\hbar^2(q^2 + \mu^2)$. (b) $\mu \to 0$: $|f|^2 =
(2mg/\hbar^2q^2)^2$; with $q = 2k\sin(\theta/2)$ and $E =
\hbar^2k^2/2m$ this is Rutherford’s formula. (c) The $\theta \to
0$ divergence integrates to infinity: every passing particle is deflected a little by the infinite-range force; in matter, atomic electrons screen the nucleus beyond $\sim10^{-10}\,\mathrm{m}$ and cut the divergence. (d) $\hbar/m_\pi c = 197.3/140 = 1.4\,\mathrm{fm}$ — the nuclear force’s reach, predicted from the pion’s mass.

**Exercise 15.6 ★★.**

The quantum hard sphere (radius $a$), $s$ wave: outside, $u_0 =
\sin(kr + \delta_0)$ with $u_0(a) = 0$. (a) Show $\delta_0 = -ka$. (b) Show $\sigma \to 4\pi a^2$ as $k \to 0$: *four* times the geometric shadow. (c) Give the wave explanation of the factor (diffraction has no sharp shadow at long wavelength). (d) At high energy the answer tends to $2\pi a^2$ — still twice geometric (shadow diffraction): why never simply $\pi a^2$?

**Solution of Exercise 15.6.**

(a) $\sin(ka + \delta_0) = 0$ with the smallest choice: $\delta_0
= -ka$. (b) $\sigma = (4\pi/k^2)\sin^2(ka) \to 4\pi a^2$. (c) At $\lambda \gg a$ the wave feels the sphere as a point defect and rebuilds itself by diffraction all around it — “shadow” is a short-wavelength concept, and the wave answer counts the full sphere surface $4\pi a^2$. (d) Even at short wavelengths, removing a disc of wave requires diffractive filling-in — the shadow itself scatters ($\pi a^2$ of blocking $+$ $\pi a^2$ of shadow diffraction).

**Exercise 15.7 ★★.**

The twenty barns of hydrogen. Slow unpolarised neutrons on protons sample the triplet channel with weight $3/4$ and the singlet with $1/4$: $\sigma = \pi(3a_{\text{t}}^2 +
a_{\text{s}}^2)$. (a) Evaluate with $a_{\text{t}} = 5.4\,\mathrm{fm}$, $a_{\text{s}} = -23.7\,\mathrm{fm}$ and compare with the measured $20.4\,\mathrm{b}$. (b) Which channel dominates, despite its smaller statistical weight? (c) Relate $a_{\text{t}}$’s size and sign to the deuteron’s near-zero binding; what does $a_{\text{s}} < 0$ say about the singlet system? (d) Why is water such an effective moderator of reactor neutrons — and why does this [cross-section](#def-b3-scattering-theory-cross-section) matter for it?

**Solution of Exercise 15.7.**

(a) $\pi(3 \times 5.4^2 + 23.7^2)\,\mathrm{fm}^{2} = \pi \times 649
\,\mathrm{fm}^{2} = 20.4\,\mathrm{b}$ — the measured value. (b) The singlet: its huge $|a_{\text{s}}|$ outweighs its $1/4$ weight. (c) $a_{\text{t}}$ large and positive: a real bound state (the deuteron) barely below threshold; $a_{\text{s}}$ large and *negative*: a “virtual” level barely above — almost a second deuteron that nature declined to bind. (d) Hydrogen’s nuclei are the best momentum-matched partners for neutrons (equal masses), and twenty barns of elastic scattering per proton makes water a superbly compact moderator.

**Exercise 15.8 ★★.**

Ramsauer–Townsend. Model the argon atom, for an incoming electron, as an attractive square well of range $R =
0.2\,\mathrm{nm}$. (a) Inside, the wave number is $K =
\sqrt{2m(E + V_0)}/\hbar$: matching can make the outside wave emerge with $\delta_0 = \pi$ exactly — what is then the $s$-wave [cross-section](#def-b3-scattering-theory-cross-section)? (b) Why does the atom become nearly invisible at that energy although the well is strong? (c) Estimate the $V_0$ that puts the transparency near $E = 1\,\mathrm{eV}$ (make the well hold half a wavelength: $KR \approx \pi$). (d) Why do helium and neon *not* show the effect at comparable energies (their wells are too shallow for $\delta_0$ to reach $\pi$) — and what does the effect’s mere existence prove about matter waves?

**Solution of Exercise 15.8.**

(a) $\sigma_0 = (4\pi/k^2)\sin^2\pi = 0$: the $s$ wave exits exactly as if no atom were there. (b) The wave is strongly distorted *inside*, but emerges with an integer number of extra half-waves: no observable phase offset, no scattering — destructive interference of the scattered wavelets. (c) $KR
\approx \pi$ with $E \ll V_0$: $V_0 \approx \pi^2\hbar^2/2mR^2
\approx 9\,\mathrm{eV}$ — an atomic-scale well. (d) Their shallower wells never wind the phase through $\pi$ at eV energies. The effect is un-mimickable classically: a transparent window in a strong attraction exists only for waves.

**Exercise 15.9 ★★.**

For a pure $s$-wave amplitude $f = \eu^{\iu\delta_0}\sin\delta_0/
k$: (a) compute $\sigma$; (b) compute $(4\pi/k)\operatorname{Im}
f(0)$ and verify the optical theorem; (c) show the maximal $s$-wave [cross-section](#def-b3-scattering-theory-cross-section) is $4\pi/k^2$ (the unitarity limit) and evaluate it for thermal neutrons ($k = 3.1 \times 10^{10}\,\mathrm{m}^{-1}$) in barns; (d) why can a resonance never push $\sigma_0$ beyond that ceiling, however strong the interaction?

**Solution of Exercise 15.9.**

(a) $\sigma = 4\pi\sin^2\delta_0/k^2$. (b) $\operatorname{Im}f(0)
= \sin^2\delta_0/k$: $(4\pi/k)\operatorname{Im}f(0) = \sigma$ — verified. (c) $\sin^2\delta_0 \le 1$: ceiling $4\pi/k^2 \approx
1 \times 10^{-20}\,\mathrm{m}^{2} = 10^{8}$ barns for thermal neutrons — room above even the most monstrous absorbers (xenon-135’s millions of barns). (d) The ceiling is unitarity: a wave cannot remove more flux than interference allows; strength saturates the sine, never exceeds it.

**Exercise 15.10 ★★★.**

A neutron resonance. Uranium-238 shows a famous resonance for neutrons at $E_{\text{R}} = 6.67\,\mathrm{eV}$ with total width $\Gamma = 27\,\mathrm{meV}$. (a) Compute the resonance’s lifetime. (b) Compute $4\pi/k^2$ at that energy, in barns, and compare with the measured peak of $\sim2 \times 10^{4}\,\mathrm{b}$ (the difference is the branching factor between elastic and capture channels — comment). (c) The width in temperature units: why does Doppler broadening of this resonance with fuel temperature matter for reactor stability (which sign of feedback)? (d) In two sentences: how do such resonances make $^{238}$U a neutron absorber in the epithermal range, and why is that central to reactor design.

**Solution of Exercise 15.10.**

(a) $\tau = \hbar/\Gamma = 6.58 \times 10^{-16}/0.027 =
2.4 \times 10^{-14}\,\mathrm{s}$ — long on nuclear timescales: a compound nucleus that “forgets” its formation. (b) $4\pi/k^2 \approx
3.9 \times 10^{5}$ b; the observed $2.3 \times 10^{4}$ b is the ceiling times the branching fraction of the entrance channel ($\Gamma_n/\Gamma
\sim 0.06$) — resonances sell tickets by partial widths. (c) $\Gamma/k_{\text{B}} \approx 310\,\mathrm{K}$: heating the fuel Doppler-widens the resonance, catching more neutrons during slow-down — absorption grows with temperature, a prompt *negative* feedback built into uranium itself. (d) Between thermal and fast energies, $^{238}$U’s resonance forest devours neutrons; fuel geometry and moderators are designed to sneak neutrons past it, and its Doppler feedback is a pillar of reactor safety.

**Exercise 15.11 ★★★.**

Cold atoms live on one number. (a) For identical bosons the low-energy [cross-section](#def-b3-scattering-theory-cross-section) is $8\pi a^2$ (constructive exchange interference — accept): evaluate for rubidium with $a =
100\,a_0$. (b) At $T = 100\,\mathrm{nK}$, check that $ka \ll 1$ (compute $k$ from $k_{\text{B}}T = \hbar^2k^2/2m$, $m =
1.4 \times 10^{-25}\,\mathrm{kg}$): the gas genuinely forgets everything but $a$. (c) Near a “Feshbach” resonance a magnetic field drags a molecular level through zero energy and $a$ diverges and changes sign, exactly as in [Example 15.6](#ex-b3-scattering-theory-np): what becomes of the gas’s interactions at will? (d) Why is this tunability — interaction strength on a dial — a physicist’s dream instrument (name one use: making molecules, simulating strongly-coupled matter, collapsing condensates)?

**Solution of Exercise 15.11.**

(a) $a = 5.3\,\mathrm{nm}$: $\sigma = 8\pi a^2 = 7 \times 10^{-16}\,\mathrm{m}^{2}$. (b) $k = \sqrt{2mk_{\text{B}}T}/\hbar \approx 6 \times 10^{6}\,\mathrm{m}^{-1}$: $ka \approx 0.03 \ll 1$. (c) The gas can be dialled from ideal ($a \approx 0$) through strongly repulsive to attractive and unstable — interactions as an experimental knob. (d) Sweeping across the resonance pairs atoms into molecules; at $a \to
\infty$ the gas becomes as strongly coupled as neutron-star matter in tabletop form — quantum simulation by [scattering length](#thm-b3-scattering-theory-partial).

**Exercise 15.12 ★★★.**

Form factors: weighing charge clouds. For an extended charge density $\rho(\vect r)$, the Born amplitude multiplies the point answer by $F(q) = \int\rho(\vect r)\eu^{\iu\vect q\cdot\vect r}
\dd^3r/Q$. (a) Show $F(0) = 1$. (b) Expand for small $q$: $F
\approx 1 - q^2\langle r^2\rangle/6$: measuring the low-$q$ fall-off weighs $\langle r^2\rangle$. (c) Electron–proton scattering fits $\sqrt{\langle r^2\rangle} \approx 0.84\,\mathrm{fm}$: which independent measurement from [Problem 11.1](https://one-course.com/books/physics/5/en/chapter/11-the-hydrogen-atom#pb-b3-hydrogen-atom-1) (muonic hydrogen) cross-checks it, and why was their historical disagreement (“the proton radius puzzle”) taken so seriously? (d) At $q \gg 1/r_{\text{p}}$ the elastic form factor collapses but hard scattering persists off *point-like constituents*: state in one sentence what 1968’s deep-inelastic version of Rutherford’s experiment found inside the proton.

**Solution of Exercise 15.12.**

(a) $F(0) = \int\rho/Q = 1$. (b) Expand the exponential; the linear term averages to zero, the quadratic gives $-q^2\langle r^2\rangle/6$. (c) Muonic hydrogen’s Lamb shift ([Problem 11.1](https://one-course.com/books/physics/5/en/chapter/11-the-hydrogen-atom#pb-b3-hydrogen-atom-1)) weighs the same $\langle r^2\rangle$ by a wholly different method; their disagreement at the $4\%$ level (0.877 versus $0.841\,\mathrm{fm}$) implied either subtle systematics or new physics — hence a decade of re-measurement (systematics, mostly, as it settled). (d) The proton is a soft cloud with hard grains inside: quarks — Rutherford’s argument, one level down.

## 15.6 Problem: The flash counters who found the nucleus

**Problem 15.1.**

Weekend problem — Rutherford scattering, from geometry to quarks

In 1909 Geiger and Marsden sat in the dark counting scintillation flashes: alpha particles fired through gold foil, a few bouncing nearly backwards — “as if a shell had rebounded off tissue paper”. Rutherford’s 1911 analysis of those counts created the nuclear atom. This problem rebuilds it. Data: alpha energy $E =
5.5\,\mathrm{MeV}$, charge $z = 2$; gold $Z = 79$, foil thickness $t
= 0.4\,\text{µ}\mathrm{m}$, $n = 5.9 \times 10^{28}\,\mathrm{atoms}/\mathrm{m}^{3}$; $k_C =
1/4\pi\varepsilon_0 = 8.99 \times 10^{9}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{C}^{2}$.

**Part I — One alpha, one nucleus.**

1. In the plum-pudding picture (charge spread over the atom, $10^{-10}\,\mathrm{m}$ ), estimate the maximum deflection a single atom could give a $5.5\,\mathrm{MeV}$ alpha (field of a smeared sphere: $\theta \sim$ force $\times$ time / momentum $\sim 2Zze^2k_C/(E\cdot10^{-10}\,\mathrm{m})$ in radians). Could any accumulation of such kicks send alphas backwards?
2. Now let the charge be a point. Write the distance of closest approach $d_0$ for a head-on collision (all kinetic energy into Coulomb energy) and evaluate it.
3. What does a measurable rate of near-backward scattering therefore *immediately* imply about how concentrated the atom’s positive charge is?
4. For a general impact parameter $b$ , the classical orbit gives $\tan(\theta/2) = d_0/2b$ (admitted — the hyperbola of the Year 1 volume’s gravity chapter, with repulsion): check its limits at $b \to 0$ and $b \to  \infty$ .
5. Compute the impact parameter that scatters an alpha by more than $90^\circ$ , and the fraction of alphas passing the foil that come within it of some nucleus ( $n t\,\pi  b^2$ ).
6. Geiger and Marsden found about 1 in 8000 alphas deflected beyond $90^\circ$ : compare.

**Part II — The [cross-section](#def-b3-scattering-theory-cross-section).**

7. Alphas with impact parameter in $[b, b + \dd b]$ scatter into $[\theta, \theta + \dd\theta]$: from $\dd\sigma =  2\pi b\,|\dd b|$ and the orbit relation, derive $$\frac{\dd\sigma}{\dd\Omega} =  \Big(\frac{d_0}{4}\Big)^2\frac{1}{\sin^4(\theta/2)} .$$
8. Verify this matches the Born/quantum answer of [Example 15.4](#ex-b3-scattering-theory-rutherford) (rewrite $d_0$ in terms of $E$ ).
9. Evaluate $\dd\sigma/\dd\Omega$ at $\theta = 30^\circ$ , $60^\circ$ , $120^\circ$ (in barns per steradian).
10. Geiger and Marsden’s 1913 counts at fixed geometry scale as $1{:}\,\sin^{-4}(\theta/2)$ : compute the predicted count ratio between $30^\circ$ and $120^\circ$ and note that their data followed such ratios over five orders of magnitude of rate.
11. Why does the same formula’s $Z^2$ let the experiment *weigh* the nuclear charge — and how did such fits help pin $Z$ (gold: 79) as the atomic number?
12. The $1/E^2$ : what happens to the whole angular pattern when the alpha energy is doubled?

**Part III — The size of the nucleus.**

13. Closest approach at angle $\theta$ : $d(\theta) =  \tfrac{d_0}2[1 + 1/\sin(\theta/2)]$ (admitted). Evaluate for $\theta = 180^\circ$ and $60{}^{\circ}$ .
14. As long as $d(\theta)$ exceeds the nuclear radius, the point-charge formula holds: what upper bound on the gold nucleus’s size did Rutherford’s verified $180^\circ$ points establish?
15. With higher-energy projectiles the counts at large angle *fall below* Rutherford: what does the departure signal (which new force, at which distance)?
16. Modern electron scattering gives nuclear radii $R  \approx r_0A^{1/3}$ with $r_0 = 1.2\,\mathrm{fm}$ : evaluate for gold ( $A = 197$ ) and compare with your bound.
17. The $A^{1/3}$ : what does it say about nuclear matter’s density (constant? growing?) — a fact [Chapter 25](https://one-course.com/books/physics/5/en/chapter/25-nuclear-physics#ch-b3-nuclear-physics) will build on.
18. Why did the (charge-blind) plum-pudding model die from *this* experiment rather than from spectroscopy?

**Part IV — Rutherford’s grandchildren.**

19. List the exact translation table between 1911 and 1968: alpha $\to$ electron beam, gold atom $\to$ proton, nucleus $\to$ quarks — what played the role of the “unexpected large-angle events” at SLAC?
20. Why are *electrons* the cleaner probe (what complication of alpha–nucleus scattering do they not have)?
21. In Born language: elastic scattering at high $q$ measures the form factor’s collapse (a soft cloud), while deep-inelastic rates stayed large and quasi-point-like: state the conclusion drawn.
22. A century’s arc in three rows: give (probe energy, distance resolved) for 1911 alphas ( $\sim10^{-14}\,\mathrm{m}$ ), 1968 SLAC ( $20\,\mathrm{GeV}$ , $\sim10^{-16}\,\mathrm{m}$ ), and the LHC (TeV scale, $\sim10^{-19}\,\mathrm{m}$ ) — using the rule “resolution $\sim \hbar c/(qc)$ ”.
23. Neutrons, uncharged, scatter only off nuclei (and off magnetic moments): name two things neutron scattering therefore maps in materials that X-rays see poorly (light atoms such as hydrogen; magnetic order).
24. Discovery [cross-sections](#def-b3-scattering-theory-cross-section) at the LHC are measured in *femtobarns* ( $10^{-43}\,\mathrm{m}^{2}$ ): from the barn to the femtobarn is fifteen orders of magnitude — what does that say about how rare the interesting collisions are, and why luminosity (delivered flux) is as precious as energy?
25. Summarise the named result: a $\sin^{-4}(\theta/2)$ law, verified flash by flash, put $99.98\%$ of the atom’s mass into a volume $10^{-14}$ of its size — and the same experiment, repeated ever harder, has never stopped finding the next layer.

**Solution of Problem 15.1.**

**1.** $\theta \sim 8 \times 10^{-4}$ rad per atom; even a random walk through $10^4$ atomic layers accumulates only a few degrees: backward scattering is impossible in the pudding. **2.** $E = k_CzZe^2/d_0$: $d_0 = k_CzZe^2/E =
4.1 \times 10^{-14}\,\mathrm{m}$ — forty femtometres. **3.** That the full positive charge sits inside $\sim4 \times 10^{-14}\,\mathrm{m}$: ten thousand times smaller than the atom. **4.** $b \to 0$: $\theta \to 180^\circ$; $b \to \infty$: $\theta \to 0$ — head-on rebounds, distant grazes. **5.** $\theta > 90^\circ$ needs $b < d_0/2 =
2.1 \times 10^{-14}\,\mathrm{m}$: fraction $nt\,\pi b^2 \approx 3 \times 10^{-5}$ — one alpha in thirty thousand for this foil. **6.** The same order as Geiger and Marsden’s one in eight thousand (their foils were thicker): the “impossible” rebounds arrive exactly as often as a point nucleus demands. **7.** $\dd\sigma = 2\pi b|\dd b|$ with $b =
(d_0/2)\cot(\theta/2)$: $|\dd b/\dd\theta| = (d_0/4)/\sin^2(
\theta/2)$, and $\dd\Omega = 2\pi\sin\theta\,\dd\theta$; assembling, the $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$ cancellations leave $(d_0/4)^2\sin^{-4}(\theta/2)$. **8.** $d_0/4 = k_CzZe^2/4E = Q_1Q_2/16\pi\varepsilon_0E$: the Born result — the $1/r$ coincidence. **9.** $(d_0/4)^2 = 1.06\,\mathrm{b}/\mathrm{sr}$: $236$, $17$ and $1.9\,\mathrm{b}/\mathrm{sr}$ at $30^\circ$, $60^\circ$, $120^\circ$. **10.** $(\sin60^\circ/\sin15^\circ)^4 \approx 125$: their counts tracked such ratios across five decades of rate — the law, not a trend. **11.** The rate scales as $Z^2$ at fixed geometry: comparing foils calibrates the nuclear charge itself, feeding the identification of $Z$ with the atomic number. **12.** Every rate drops fourfold; the angular *shape* is untouched — a clean experimental signature of $1/E^2$. **13.** $d(180^\circ) = d_0 = 41\,\mathrm{fm}$; $d(60^\circ) =
(d_0/2)(1 + 2) = 62\,\mathrm{fm}$. **14.** The formula held at the largest angles: the gold nucleus is smaller than $\sim4 \times 10^{-14}\,\mathrm{m}$. **15.** The projectile begins to touch the nucleus: the short-range *strong* force (and absorption) sets in at femtometre distances — the departure measures the nuclear edge. **16.** $R = 1.2 \times 197^{1/3} = 7.0\,\mathrm{fm}$ — comfortably inside Rutherford’s bound. **17.** Volume $\propto A$: nuclear matter has a fixed density ($\sim2 \times 10^{17}\,\mathrm{kg}/\mathrm{m}^{3}$) — nuclei are droplets of an incompressible liquid, the starting picture of [Chapter 25](https://one-course.com/books/physics/5/en/chapter/25-nuclear-physics#ch-b3-nuclear-physics). **18.** Spectroscopy interrogated the electrons; only a projectile that penetrates the atom could testify where the *positive* charge and the mass sit. **19.** Electrons at $20$ GeV scattering at improbably large angles and energy losses — too many hard events for a soft proton: Rutherford’s rebound, re-enacted. **20.** Electrons feel no strong force and have no known substructure: a point probe reading charge alone, with none of the alpha’s own compositeness. **21.** The elastic form factor’s collapse says the proton is a diffuse cloud; the persistence of hard *inelastic* scattering says the cloud contains point-like constituents — quarks (partons). **22.** 1911: $\sim10^{-14}\,\mathrm{m}$; SLAC: $\hbar c/q \sim
0.2\,\mathrm{fm} = 2 \times 10^{-16}\,\mathrm{m}$; LHC: $\sim2 \times 10^{-19}\,\mathrm{m}$ — five orders of magnitude of ruler in one century. **23.** Hydrogen positions (in ice, polymers, proteins) and magnetic structures (antiferromagnets, spin spirals) — both nearly invisible to X-rays, both bread and butter for neutrons. **24.** The processes worth discovering occur once per $10^{15}$ ordinary encounters: only colossal luminosity turns femtobarns into events per year — collider design is [cross-section](#def-b3-scattering-theory-cross-section) arithmetic. **25.** A $\sin^{-4}(\theta/2)$ law, checked flash by flash, concentrated the atom’s mass into $10^{-14}$ of its volume; run at ever higher $q$, the same experiment found the nucleus’s size, then its constituents, and is still looking.