Rutherford Scattering Cross-Section

Derivation

Rutherford Scattering Cross-Section

D-273 Home PU-304 Threads force · fields · chance Depends on kepler-orbit-hyperbolic, Born Approximation and the Form Factor

Statement

For a non-relativistic point charge \(z_1 e\) scattering off a fixed point charge \(z_2 e\) through the Coulomb interaction \(V(r) = \frac{z_1 z_2 e^2}{4\pi\varepsilon_0 r}\), the differential cross-section in the centre-of-mass frame is \(\frac{d\sigma}{d\Omega} = \left(\frac{z_1 z_2 e^2}{16\pi\varepsilon_0 E}\right)^2 \csc^4\!\left(\frac{\theta}{2}\right)\), where \(E = \tfrac{1}{2}\mu v_\infty^2\) is the kinetic energy at infinity, \(\theta\) is the scattering angle, and \(\mu\) is the reduced mass. This is derived below from the classical hyperbolic orbit and shown to be reproduced identically by the first Born approximation.

Why it matters

This is the result that dismantled the plum-pudding atom. Geiger and Marsden's 1909 observation of alpha particles back-scattered through more than \(90^\circ\) is impossible for a diffuse charge distribution; the \(\csc^4(\theta/2)\) law, verified across four orders of magnitude in rate, forced Rutherford's 1911 nuclear model. The cross-section is the quantitative bridge between a microscopic force law and a macroscopic counting experiment.

It is also a rare exact meeting point of classical and quantum mechanics. The \(1/r\) potential is special: the classical orbit calculation, the first Born approximation, and the full partial-wave (Gordon) solution all yield the same closed form. That coincidence — and the divergences that expose its limits — makes Rutherford scattering the canonical worked example in every scattering-theory course.

Assumptions

Fixed, infinitely heavy target (or work in the CM frame).If the target recoils, the lab-frame angle differs from the CM angle and \(\frac{d\sigma}{d\Omega}\) must be transformed by the Jacobian \(d\Omega_{\text{cm}}/d\Omega_{\text{lab}}\); using the target mass in place of \(\mu\) then mis-scales the energy. Pure Coulomb potential, unscreened and point-like.Atomic electrons screen the nucleus at large \(r\), cutting off the forward divergence; a finite nuclear size cuts it off at large angle. Dropping "point-like" invalidates the small-\(\theta\) tail and the large-\(\theta\) tail respectively. Non-relativistic kinematics, \(v_\infty \ll c\).At relativistic energies the Mott cross-section replaces this, adding a \((1-\beta^2\sin^2(\theta/2))\) spin factor absent here. Single scattering; the detector subtends one nucleus.In a thick foil multiple small-angle deflections compound (Molière theory); the single-scatter cross-section then describes only the wide-angle tail of the observed distribution. Distinguishable projectile and target.For identical particles (e.g. \(\alpha\)–\(\alpha\), \(e^-\)–\(e^-\)) quantum exchange adds the Mott interference term \(\propto \cos[\eta \ln\tan^2(\theta/2)]\), which the classical derivation cannot see.

Derivation

\[ E = \tfrac{1}{2}\mu v_\infty^2, \qquad L = \mu v_\infty b \]

Energy and angular momentum are conserved in a central field; evaluate both far from the scatterer where \(V\to 0\) and the trajectory is a straight line at impact parameter \(b\). A

\[ \frac{1}{r} = \frac{\mu \kappa}{L^2}\bigl(e\cos\phi - 1\bigr), \qquad \kappa \equiv \frac{z_1 z_2 e^2}{4\pi\varepsilon_0} \]

This is the assumed hyperbolic Kepler orbit for a repulsive \(V=\kappa/r\); eccentricity \(e>1\) since \(E>0\). The polar angle \(\phi\) is measured from the symmetry axis (periapsis). B

\[ e = \sqrt{1 + \left(\frac{2 E L}{\mu \kappa}\right)^{2}} = \sqrt{1 + \left(\frac{\mu v_\infty^2 b}{\kappa}\right)^{2}} \]

Standard eccentricity–energy relation for the Kepler orbit; substitute \(E=\tfrac12\mu v_\infty^2\) and \(L=\mu v_\infty b\) from Step 1 to express \(e\) purely in terms of the initial conditions. B

\[ r\to\infty \;\Rightarrow\; \cos\phi_\infty = \frac{1}{e} \]

The asymptotes occur where \(1/r\to 0\) in Step 2, i.e. \(e\cos\phi_\infty = 1\). The two asymptotic directions are \(\pm\phi_\infty\). A

\[ \theta = \pi - 2\phi_\infty \quad\Longrightarrow\quad \phi_\infty = \frac{\pi}{2} - \frac{\theta}{2} \]

Geometry of the hyperbola: the scattering angle is the supplement of the angle between the two asymptotes. Hence \(\cos\phi_\infty = \sin(\theta/2)\). A

\[ \sin\frac{\theta}{2} = \frac{1}{e} \quad\Longrightarrow\quad \cot\frac{\theta}{2} = \sqrt{e^2 - 1} = \frac{\mu v_\infty^2 b}{\kappa} \]

Combine Steps 4–5 (\(\cos\phi_\infty = 1/e = \sin(\theta/2)\)); then \(\cot(\theta/2) = \cos(\theta/2)/\sin(\theta/2) = \sqrt{e^2-1}\), using \(e^2-1\) from Step 3. B

\[ b(\theta) = \frac{\kappa}{\mu v_\infty^2}\cot\frac{\theta}{2} = \frac{\kappa}{2E}\cot\frac{\theta}{2} \]

Invert Step 6 to obtain the deflection function — the one-to-one map from impact parameter to scattering angle. Large \(b\) gives small \(\theta\); head-on (\(b\to0\)) gives back-scatter (\(\theta\to\pi\)). A

\[ \frac{d\sigma}{d\Omega} = \frac{b}{\sin\theta}\left|\frac{db}{d\theta}\right| \]

Classical cross-section definition: particles entering the annulus \(2\pi b\,db\) exit into the solid-angle ring \(2\pi\sin\theta\,d\theta\); flux conservation gives this ratio. The absolute value handles the monotonically decreasing \(b(\theta)\). A

\[ \frac{db}{d\theta} = -\frac{\kappa}{2E}\cdot\frac{1}{2}\csc^2\frac{\theta}{2} \]

Differentiate Step 7, using \(\frac{d}{dx}\cot x = -\csc^2 x\) and the chain-rule factor \(\tfrac12\) from the argument \(\theta/2\). B

\[ \frac{d\sigma}{d\Omega} = \frac{1}{\sin\theta}\cdot\frac{\kappa}{2E}\cot\frac{\theta}{2}\cdot\frac{\kappa}{4E}\csc^2\frac{\theta}{2} \]

Insert Steps 7 and 9 into Step 8. Everything is now symbolic in \(\kappa\), \(E\), \(\theta\); numbers enter only in the worked examples. B

\[ \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}, \qquad \cot\frac{\theta}{2}=\frac{\cos(\theta/2)}{\sin(\theta/2)} \]

Apply the double-angle identity to \(\sin\theta\) so the \(\cos(\theta/2)\) factors cancel against \(\cot(\theta/2)\), leaving a pure power of \(\csc(\theta/2)\). B

\[ \frac{d\sigma}{d\Omega} = \frac{\kappa^2}{8E^2}\cdot\frac{\cos(\theta/2)/\sin(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\csc^2\frac{\theta}{2} = \left(\frac{\kappa}{4E}\right)^{2}\csc^4\frac{\theta}{2} \]

Cancel \(\cos(\theta/2)\), collect the remaining \(\sin^{-2}(\theta/2)\) with the explicit \(\csc^2(\theta/2)\), and gather constants. This is the Rutherford formula. A

\[ f(\theta) = -\frac{\mu}{2\pi\hbar^2}\int e^{i\mathbf{q}\cdot\mathbf{r}}\,V(r)\,d^3r = -\frac{2\mu\kappa}{\hbar^2 q^2}, \qquad q = 2k\sin\frac{\theta}{2} \]

First Born approximation with the assumed scattering amplitude. The Fourier transform of the screened Coulomb \(e^{-\lambda r}/r\) is \(4\pi/(q^2+\lambda^2)\); take \(\lambda\to0\). Momentum transfer \(q=|\mathbf{k}'-\mathbf{k}|=2k\sin(\theta/2)\) with \(\hbar k = \mu v_\infty\). C

\[ \frac{d\sigma}{d\Omega} = |f(\theta)|^2 = \frac{4\mu^2\kappa^2}{\hbar^4 q^4} = \frac{4\mu^2\kappa^2}{\hbar^4 (2k\sin(\theta/2))^4} = \left(\frac{\kappa}{4E}\right)^2 \csc^4\frac{\theta}{2} \]

Square the amplitude; substitute \(q=2k\sin(\theta/2)\) and \(\hbar^2 k^2 = 2\mu E\) so that \(\hbar^4 k^4 = 4\mu^2 E^2\). The reduced mass, \(\hbar\), and \(k\) all cancel, reproducing Step 12 exactly — no \(\hbar\) survives. C

Result

\[ \frac{d\sigma}{d\Omega} = \left(\frac{z_1 z_2 e^2}{16\pi\varepsilon_0 E}\right)^{2}\csc^4\!\left(\frac{\theta}{2}\right) = \left(\frac{\kappa}{4E}\right)^2 \frac{1}{\sin^4(\theta/2)} \]

Reading. The number of particles scattered into unit solid angle at angle \(\theta\) falls off as \(\sin^{-4}(\theta/2)\): it is enormous in the forward direction and drops steeply toward back-scatter, yet never vanishes — even \(\theta=\pi\) has a finite, non-zero rate set entirely by \(\kappa/4E\). The cross-section scales as \((z_1 z_2)^2\) (charge product squared) and as \(E^{-2}\) (faster projectiles are deflected less). Crucially, \(\hbar\) is absent: the classical orbit and the quantum Born result are numerically identical for the \(1/r\) potential.

Units check. \(\kappa = z_1 z_2 e^2/4\pi\varepsilon_0\) has units of \(\text{J}\cdot\text{m}\) (energy \(\times\) length, since \(V=\kappa/r\) is an energy). Thus \(\kappa/E\) has units of metres, \((\kappa/4E)^2\) has units of \(\text{m}^2\), and \(\csc^4(\theta/2)\) is dimensionless — so \(d\sigma/d\Omega\) is an area per steradian, as a differential cross-section must be. In SI, \(1\ \text{barn}=10^{-28}\ \text{m}^2\).

Limiting cases

Forward limit \(\theta\to0\): \(\csc^4(\theta/2)\to 16/\theta^4\to\infty\). The cross-section diverges because the unscreened \(1/r\) tail has infinite range; every impact parameter, however large, produces some deflection.
Total cross-section \(\sigma_{\text{tot}}=\int d\sigma\): \(\int \csc^4(\theta/2)\sin\theta\,d\theta\) diverges logarithmically at small \(\theta\) — the Coulomb potential has no finite total cross-section, a direct signature of its infinite range.
Back-scatter \(\theta\to\pi\): \(\csc^4(\theta/2)\to 1\), the minimum. The rate is finite and equals \((\kappa/4E)^2\) per steradian — the distance of closest approach in a head-on collision, \(d_{\min}=\kappa/E\), sets this scale.
Small-angle / high-energy \(E\to\infty\): \(d\sigma/d\Omega\propto E^{-2}\); the whole pattern collapses forward, matching the impulse (straight-line) approximation \(\theta\approx 2\kappa/(\mu v_\infty^2 b)\).
Rutherford \(z_1 z_2<0\) (attractive): the formula is unchanged — it depends only on \((z_1 z_2)^2\); classically attraction and repulsion give the same cross-section because \(b(\theta)\) has the same magnitude.

Breaks when

Small angles (screening). Below \(\theta_{\min}\sim\hbar/(p\,a)\), where \(a\) is the atomic screening radius, bound electrons neutralise the nuclear charge. The \(\csc^4\) divergence is cut off and the integrated cross-section becomes finite; the bare formula overestimates the forward rate.
Large angles (finite nuclear size). When the closest approach \(d_{\min}=\kappa/E\) becomes comparable to the nuclear radius \(R\sim 1.2\,A^{1/3}\ \text{fm}\), the projectile probes inside the charge distribution and feels the nuclear force. Measured rates fall below Rutherford — historically how nuclear radii were first extracted.
Relativistic / spin regime. For \(v_\infty\sim c\) the non-relativistic kinematics fail; electron scattering requires the Mott cross-section with its \(1-\beta^2\sin^2(\theta/2)\) factor and, for a Dirac target, a form factor.
Identical particles. For \(\alpha\)–\(\alpha\) or \(e^-\)–\(e^-\), quantum exchange symmetry produces interference between \(\theta\) and \(\pi-\theta\) trajectories; the smooth classical curve is replaced by the oscillating Mott formula.

Failure modes

Angle-halving slip: writing \(\csc^4\theta\) instead of \(\csc^4(\theta/2)\). The half-angle comes from \(\theta=\pi-2\phi_\infty\); dropping it puts the back-scatter minimum in the wrong place and doubles the forward exponent's onset.
Using the target mass instead of \(\mu\): forgetting that \(E=\tfrac12\mu v_\infty^2\) in the CM frame. For an \(\alpha\) on gold the correction is tiny, but for comparable masses it mis-scales \(\kappa/E\) badly.
Lab–CM confusion: quoting the CM \(\csc^4\) law against lab-measured angles without the Jacobian transform. Correct only when the target is effectively infinite-mass.
Factor-of-2 in \(\kappa\): writing \(\kappa=z_1 z_2 e^2/(2\pi\varepsilon_0)\) or dropping the \(4\pi\varepsilon_0\); check that \(d_{\min}=\kappa/E\) reproduces the known closest-approach formula.
Squaring error: reporting \(d\sigma/d\Omega\propto \csc^2\) by squaring the amplitude's \(q^{-2}\) but forgetting that \(q^4\) already appears, or vice versa. The Born route squares \(f\propto q^{-2}\) to get \(q^{-4}\).
Believing the total cross-section is finite: attempting \(\int d\sigma\) and getting a number by silently truncating the divergent forward integral.

Discussion

The most remarkable feature of this result is what is absent: Planck's constant. For a general potential the classical and quantum cross-sections differ, and \(\hbar\) sets the scale of the difference. The Coulomb \(1/r\) potential is the exception — its scale invariance means the only length in the problem is \(\kappa/E\), which is purely classical, so the Born, classical, and exact partial-wave calculations must all collapse onto the same \((\kappa/4E)^2\csc^4(\theta/2)\). This coincidence let Rutherford, working entirely classically in 1911, extract the correct quantum answer decades before quantum mechanics existed.

The threads of the course meet here. Force supplies the \(1/r^2\) Coulomb law and the hyperbolic orbit; fields supply the potential energy \(\kappa/r\) whose Fourier transform \(4\pi/q^2\) is the propagator squared in the Born route; chance enters through the cross-section itself — a probability per unit incident flux that a single quantum event deposits a particle in a given detector. The deflection function \(b(\theta)\) is the deterministic classical skeleton; the cross-section dresses it in the statistics of a beam.

The forward divergence is not a defect but a diagnostic. Its logarithmic character is the fingerprint of a long-range potential, and it reappears throughout physics: the infrared divergences of QED (soft-photon emission), the need for Coulomb-modified asymptotic states (the Coulomb phase \(\eta\ln\)-terms) rather than plane waves, and the Sommerfeld/Gamow factor \(2\pi\eta/(e^{2\pi\eta}-1)\) that governs fusion rates in stars all trace to the same \(1/r\) tail that makes \(\sigma_{\text{tot}}\) diverge. The exact Coulomb amplitude carries a divergent phase for exactly this reason, even though its modulus reproduces Rutherford.

Common misconceptions. (i) "The finite result at \(\theta=\pi\) means alpha particles bounce straight back off a hard sphere" — no; the nucleus is a point charge, and the finite back-rate reflects the closest-approach scale \(\kappa/E\), not a surface. (ii) "The formula proves the atom is mostly empty space" — it proves the positive charge is concentrated; the emptiness inference is separate. (iii) "Because \(\hbar\) cancels, quantum mechanics is irrelevant here" — it cancels only in the modulus for \(1/r\); the phase, identical-particle interference, and screening are all irreducibly quantum.

Worked examples

Alpha particle (\(z_1=2\)) on gold (\(z_2=79\)), \(E=5.0\ \text{MeV}\); find \(d\sigma/d\Omega\) at \(\theta=60^\circ\) and the distance of closest approach in a head-on hit.

Symbolic first: \(d\sigma/d\Omega=(\kappa/4E)^2\csc^4(\theta/2)\) with \(\kappa=z_1 z_2 e^2/4\pi\varepsilon_0\); \(d_{\min}=\kappa/E\). A

\[ \frac{\kappa}{4E}=\frac{z_1 z_2 e^2}{16\pi\varepsilon_0 E} = \frac{z_1 z_2 (1.44\ \text{MeV·fm})}{4E} \]

Use the convenient constant \(e^2/4\pi\varepsilon_0 = 1.44\ \text{MeV·fm}\), so \(\kappa = z_1 z_2\times 1.44\ \text{MeV·fm} = 2\cdot79\cdot1.44 = 227.5\ \text{MeV·fm}\). A

\[ \frac{\kappa}{4E}=\frac{227.5\ \text{MeV·fm}}{4\times 5.0\ \text{MeV}} = 11.4\ \text{fm} \]

Energy cancels in MeV; result is a length. A

\[ \csc^4(30^\circ) = \left(\frac{1}{0.5}\right)^4 = 16 \]

\(\sin(60^\circ/2)=\sin 30^\circ = 0.5\). A

\[ \frac{d\sigma}{d\Omega} = (11.4\ \text{fm})^2\times 16 \approx 2.07\times10^3\ \text{fm}^2/\text{sr} = 20.7\ \text{b/sr} \]

Reading. About 21 barn per steradian at \(60^\circ\). Head-on closest approach \(d_{\min}=\kappa/E = 227.5/5.0 = 45.5\ \text{fm}\) — far larger than the gold nuclear radius \(R\approx 1.2\cdot197^{1/3}\approx 7\ \text{fm}\), confirming the \(5\ \text{MeV}\) alpha never touches the nucleus, so pure Rutherford holds.

A proton (\(z_1=1\)) scatters off a gold nucleus (\(z_2=79\)) at \(E=10\ \text{MeV}\). Find the ratio of scattered intensity at \(\theta=30^\circ\) to that at \(\theta=90^\circ\), and the impact parameter that produces \(90^\circ\).

The \((\kappa/4E)^2\) prefactor cancels in the ratio: \(\dfrac{d\sigma/d\Omega|_{30}}{d\sigma/d\Omega|_{90}} = \dfrac{\csc^4 15^\circ}{\csc^4 45^\circ}\). For \(b\): \(b=\frac{\kappa}{2E}\cot(\theta/2)\). A

\[ \frac{\csc^4 15^\circ}{\csc^4 45^\circ} = \left(\frac{\sin 45^\circ}{\sin 15^\circ}\right)^4 = \left(\frac{0.7071}{0.2588}\right)^4 \]

Ratio of cross-sections is the inverse fourth power of the sine ratio. A

\[ \left(2.732\right)^4 = 55.7 \]

\(2.732^2=7.46\), squared again \(=55.7\). The forward-peaking is dramatic even over this modest angular range. A

\[ b(90^\circ)=\frac{\kappa}{2E}\cot 45^\circ = \frac{79\times1.44\ \text{MeV·fm}}{2\times10\ \text{MeV}}\times 1 \]

\(\kappa = 1\cdot79\cdot1.44 = 113.8\ \text{MeV·fm}\); \(\cot 45^\circ = 1\). A

\[ \frac{I(30^\circ)}{I(90^\circ)} \approx 56, \qquad b(90^\circ) = 5.7\ \text{fm} \]

Reading. The \(30^\circ\) rate is about 56 times the \(90^\circ\) rate — the steep forward bias that made small-angle counting the hard part of the Geiger–Marsden experiment. A \(90^\circ\) deflection requires the proton to pass within \(5.7\ \text{fm}\) of the nucleus, comparable to \(R\approx7\ \text{fm}\); at this energy a proton would begin to feel nuclear forces, so real data at large \(\theta\) already deviate from Rutherford.

Problems

(A) An alpha particle (\(z_1=2\)) scatters off silver (\(z_2=47\)) at \(E=7.7\ \text{MeV}\). Compute \(d\sigma/d\Omega\) at \(\theta=90^\circ\) in barn/sr.
Solution
\(\kappa = 2\cdot47\cdot1.44 = 135.4\ \text{MeV·fm}\); \(\kappa/4E = 135.4/(4\cdot7.7) = 4.39\ \text{fm}\). \(\csc^4 45^\circ = (1/0.7071)^4 = 4\). \(d\sigma/d\Omega = (4.39)^2\times 4 = 77.2\ \text{fm}^2/\text{sr} = 0.77\ \text{b/sr}\).
(A) For the alpha–gold case (\(\kappa/2E\) with \(z_1 z_2=158\), \(E=5.0\ \text{MeV}\)), find the impact parameter that produces a \(10^\circ\) deflection.
Solution
\(\kappa = 158\cdot1.44 = 227.5\ \text{MeV·fm}\); \(\kappa/2E = 227.5/10 = 22.75\ \text{fm}\). \(\cot 5^\circ = 11.43\). \(b = 22.75\times 11.43 = 260\ \text{fm}\). The large \(b\) confirms small-angle scattering samples distant, near-grazing trajectories.
(B) Show that the fraction of a monoenergetic beam scattered through angles greater than \(\theta_0\) by a single nucleus corresponds to an effective area \(\sigma(>\theta_0)=\pi b(\theta_0)^2\), and evaluate it for the case of Problem 2's proton (\(\kappa=113.8\ \text{MeV·fm}\), \(E=10\ \text{MeV}\)) with \(\theta_0=90^\circ\).
Solution
All particles with \(b<b(\theta_0)\) scatter by more than \(\theta_0\) because \(b(\theta)\) decreases monotonically; the corresponding cross-sectional area is \(\pi b(\theta_0)^2\). With \(b(90^\circ)=\frac{\kappa}{2E}\cot45^\circ = \frac{113.8}{20} = 5.69\ \text{fm}\): \(\sigma(>90^\circ)=\pi(5.69)^2 = 101.7\ \text{fm}^2 = 1.02\ \text{b}\).
(B) Two experiments use the same target and geometry. Beam 1 is \(6\ \text{MeV}\) alphas; beam 2 is \(3\ \text{MeV}\) alphas. At fixed \(\theta\), what is the ratio of their differential cross-sections?
Solution
\(d\sigma/d\Omega\propto E^{-2}\) at fixed \(\theta\) and fixed charges. Ratio \(=\left(\frac{E_2}{E_1}\right)^{-2}=\left(\frac{3}{6}\right)^{-2}\)... carefully: \(\frac{(d\sigma/d\Omega)_1}{(d\sigma/d\Omega)_2}=\left(\frac{E_2}{E_1}\right)^2=\left(\frac{3}{6}\right)^2=\frac14\). The lower-energy beam 2 scatters four times more strongly, since slower particles are deflected more.
(C) Starting from the Born amplitude \(f(\theta)=-\frac{2\mu\kappa}{\hbar^2 q^2}\) with \(q=2k\sin(\theta/2)\) and \(\hbar^2 k^2 = 2\mu E\), derive the Rutherford formula and confirm explicitly that \(\hbar\) and \(\mu\) cancel. Then state one physical quantity that the classical derivation cannot reproduce.
Solution
\(\frac{d\sigma}{d\Omega}=|f|^2 = \frac{4\mu^2\kappa^2}{\hbar^4 q^4}=\frac{4\mu^2\kappa^2}{\hbar^4\,16k^4\sin^4(\theta/2)}\). Use \(\hbar^2 k^2 = 2\mu E\Rightarrow \hbar^4 k^4 = 4\mu^2 E^2\): \(=\frac{4\mu^2\kappa^2}{16\cdot 4\mu^2 E^2\sin^4(\theta/2)}=\frac{\kappa^2}{16 E^2}\csc^4(\theta/2)=\left(\frac{\kappa}{4E}\right)^2\csc^4(\theta/2)\). Both \(\mu^2\) and \(\hbar^4\) cancel identically. What the classical route misses: the Coulomb scattering phase \(\arg f\propto -\eta\ln\sin^2(\theta/2)\) (with \(\eta=\kappa/\hbar v\)), which is invisible to \(|f|^2\) but governs interference for identical particles (Mott scattering) and is genuinely quantum.