Rutherford Scattering

In 1909, in a basement laboratory in Manchester, Hans Geiger and a twenty-year-old undergraduate named Ernest Marsden sat in the dark for hours, counting faint green flashes on a zinc-sulfide screen. They were firing alpha particles at thin metal foils, expecting them to punch straight through — and mostly they did. But roughly one alpha in eight thousand came back off a platinum reflector on the same side it went in. Rutherford later said it was "as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you." The accepted atom of the day — J. J. Thomson's diffuse "plum pudding" of smeared-out positive charge — could deflect a fast alpha by a fraction of a degree at most. Something inside the atom had to be tiny, massive, and ferociously charged. Rutherford's 1911 answer was a triumph of pure classical mechanics. Put all the positive charge in a point nucleus, let the alpha swing past it on the inverse-square Coulomb force, and the trajectory is a hyperbola — the open-orbit twin of the Kepler ellipse. Working out how the deflection angle depends on the aiming offset, and converting that into a measurable angular distribution, gives the celebrated $\sin^{-4}(\Theta/2)$ law. Geiger and Marsden then confirmed every predicted scaling — angle, energy, charge, foil thickness — and the nuclear atom was born. This article is that calculation, done honestly: the hyperbolic orbit, the deflection law $\tan(\Theta/2)=\kappa/2Eb$ , the cross-section, the distance of closest approach that first bounded the size of the nucleus, and the strange quantum accident that makes this classical result exactly right.

Key takeaways

Rutherford scattering is the deflection of a charged particle by the inverse-square Coulomb force — the unbound, hyperbolic branch of the Kepler problem.
One clean relation controls everything: the deflection angle $\Theta$ and impact parameter $b$ obey $\tan(\Theta/2) = \kappa/(2Eb)$ , with $\kappa = q_1q_2/4\pi\varepsilon_0$ .
Flux conservation converts $b(\Theta)$ into the differential cross-section $\dfrac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \left(\dfrac{\kappa}{4E}\right)^2 \dfrac{1}{\sin^4(\Theta/2)}$ — steeply forward-peaked, with a finite floor at $\Theta = 180°$ .
The $E^{-2}$ , $Z^2$ , and $\sin^{-4}(\Theta/2)$ scalings were each verified by Geiger and Marsden (1913); the fit is what established the nuclear atom.
The distance of closest approach, $r_{\min} = \kappa/E$ head-on, gave the first upper bound on nuclear size: about $4\times10^{-14}$ m for MeV alphas on gold — ten thousand times smaller than the atom.
The total cross-section diverges, because the infinite-range $1/r$ tail deflects every trajectory a little; real targets are rescued by electron screening.
By a famous accident special to $1/r$ , the exact quantum cross-section equals the classical one — Rutherford's arithmetic survives quantum mechanics untouched.
Key idea: compute the scattering angle as a function of impact parameter from the hyperbolic Kepler orbit, then translate the flux-conservation relation $\mathrm{d}\sigma = 2\pi b\,|\mathrm{d}b|$ into $\mathrm{d}\sigma/\mathrm{d}\Omega = (b/\sin\Theta)|\mathrm{d}b/\mathrm{d}\Theta|$ , giving the $\sin^{-4}(\Theta/2)$ law.

Where this lives today

Rutherford's formula never retired; it industrialized. Rutherford backscattering spectrometry (RBS) fires MeV helium ions at a sample and reads the energy spectrum of the backscattered ions: the $\kappa^2 \propto Z^2$ dependence identifies which elements are present, kinematics reveals the depth at which each scattering happened, and equation (6) — used as an absolute, standards-free cross-section — turns count rates into atomic concentrations. It remains a workhorse of 2026-era semiconductor metrology, where fabs use ion-beam analysis to certify the stoichiometry and thickness of the nanometre-scale films in gate stacks and EUV-mirror multilayers. The same physics rides on other planets: the alpha-particle X-ray spectrometers on Mars rovers put a curium alpha source against a rock and decode the backscattered alphas and induced X-rays into a chemical assay. The deeper legacy is methodological. Probing structure by firing a beam and reading an angular distribution against a $1/\sin^4$ baseline is the template of subatomic physics: deviations from Rutherford at high energy measured nuclear radii in the 1950s; the analogous deep-inelastic electron-scattering experiments at SLAC in the late 1960s found hard scattering centres inside the proton — quarks — in a conscious replay of Geiger and Marsden; and today's LHC analyses still publish differential cross-sections $\mathrm{d}\sigma/\mathrm{d}\Omega$ whose Coulomb-like tails are computed from the same $\kappa^2/16E^2$ skeleton, dressed in quantum field theory. Every collider physicist is running Rutherford's experiment at higher energy.

The physical picture

Fire a uniform, parallel beam of alpha particles at a single gold nucleus and every projectile is labeled by one number: its impact parameter $b$ , the sideways offset between its initial straight line and the line through the nucleus. That one number decides its fate. A particle aimed 110 fm off-center feels only the distant tail of the Coulomb repulsion and ambles off at 21°. Aim 30 fm off and the deflection is 69°. At $b = 20.7$ fm (for a 5.5 MeV alpha on gold) the particle leaves at a right angle to how it came in. And a nearly head-on shot at $b = 2$ fm climbs almost the entire Coulomb hill, stops, and is hurled back the way it came.

Figure

One nucleus, many aims — exact trajectories of 5.5 MeV alphas on gold, from a grazing 110 fm down to a nearly head-on 2 fm

Three features of this picture carry all the physics. First, the mapping $b \to \Theta$ is monotonic: closer aim, bigger kick, with no exceptions — a special property of the bare $1/r$ potential. Second, the geometry is rigid: the hyperbola is symmetric about the line to closest approach, so the whole deflection is fixed by a single conic parameter, the eccentricity $e$ , through the tidy relation $\sin(\Theta/2) = 1/e$ . Third, and decisively for the history of physics, nothing here requires contact. The alpha that bounces straight back never touches anything — it is turned around by the electric field alone, at a closest approach of about 41 fm for a 5.5 MeV alpha, several times the actual radius of the gold nucleus. Because almost all of a beam passes at large $b$ , where the deflection is tiny, the scattered intensity is enormously concentrated in the forward direction — and falls by more than five orders of magnitude between 10° and 180°:

Figure

The Rutherford angular distribution for 5.5 MeV alphas on gold — five decades of fall-off, with a finite floor at 180 degrees

That never-zero floor at large angles is the discovery. Thomson's smeared-out atom predicts essentially zero backscattering — a fast alpha crossing a diffuse charge cloud picks up thousands of feeble, randomly-directed nudges of a hundredth of a degree each, and the odds of them conspiring to 90° or more are astronomically small. A point nucleus, by contrast, guarantees a hard core of large-angle events with a definite, calculable rate. Geiger and Marsden measured that rate, and it obeyed Rutherford's formula in every variable. One more scaling matters: the prefactor $(\kappa/4E)^2$ means the whole pattern shrinks as $1/E^2$ — faster alphas are harder to turn — and that too was checked, by slowing the beam with mica sheets.

From aim to angle: the deflection law

Set up the problem cleanly. A projectile of charge $q_1$ approaches a fixed target of charge $q_2$ from infinite distance with speed $v_\infty$ and impact parameter $b$ . (For a real two-body problem, work in the CM frame with the reduced mass $\mu$ ; for an alpha on gold, $\mu = 0.98\,m_\alpha$ , so the correction is only 2%.) The interaction is the Coulomb potential, which we write with a single strength constant so that every later formula stays compact:

V(r) = \frac{\kappa}{r}, \qquad \kappa = \frac{q_1 q_2}{4\pi\varepsilon_0} \qquad (1)

where: $\kappa > 0$ for like charges (repulsive) and $\kappa < 0$ for attraction. For an alpha particle on a gold nucleus, $\kappa = 2 \times 79 \times 1.44\ \mathrm{MeV\,fm} = 227.5\ \mathrm{MeV\,fm}$ — a number worth memorizing in this unit system, because dividing it by an energy immediately gives you a length scale. Two conserved quantities are fixed at infinity, before the force has done anything, and they control the entire encounter:

E = \tfrac{1}{2}\mu v_\infty^2 > 0, \qquad L = \mu v_\infty b \qquad (2)

where: $E$ — the total energy, positive because the particle is unbound; $L$ — the angular momentum about the scattering center. Note what equation (2) is really saying: specifying the beam energy and the aim ( $E$ , $b$ ) is equivalent to specifying the two dynamical invariants ( $E$ , $L$ ). Nothing else about the trajectory is free. Because $E>0$ , the orbit is a hyperbola — the same conic-section family as the Kepler ellipse, continued to positive energy — and the scattering angle is pure geometry: it is the angle between the two asymptotes of that hyperbola. The result, derived below, is the central formula of the subject:

\tan\!\left(\frac{\Theta}{2}\right) = \frac{\kappa}{2Eb} \qquad (3)

where: $\Theta$ — the deflection angle in the CM frame. Read it physically: the deflection grows when the force is strong ( $\kappa$ large), the projectile slow ( $E$ small), or the aim close ( $b$ small). Head-on ( $b\to0$ ) gives $\Theta\to180°$ ; a grazing pass ( $b\to\infty$ ) gives $\Theta\to0$ , but never exactly zero — the infinite range of the Coulomb tail touches every trajectory, a fact that will come back to haunt the total cross-section.

Figure

The scattering geometry for the 90-degree trajectory — impact parameter, closest approach, and the angle between asymptotes

Derivation AThe deflection angle from the hyperbolic orbit

We are deriving the deflection law (3), $\tan(\Theta/2) = \kappa/2Eb$ , from the shape of the Coulomb orbit. The strategy is the standard central-force route: write Binet's orbit equation for $u = 1/r$ , solve it (a driven harmonic oscillator in the angle variable), locate the asymptotes where $r \to \infty$ , and then convert the asymptote geometry into the deflection using the orbit's symmetry about closest approach. The step to watch is the geometric one: $\Theta$ is not the asymptote angle $\phi_\infty$ itself but its supplement partner, $\Theta = \pi - 2\phi_\infty$ — confusing the two is the single most common error in this calculation.

Step 1. Write Binet's equation for the repulsive inverse-square force $F = +\kappa/r^2$ (radially outward for $\kappa > 0$ ):

\frac{\mathrm{d}^2 u}{\mathrm{d}\phi^2} + u = -\frac{\mu\kappa}{L^2}, \qquad u \equiv \frac{1}{r}

Why: for any central force, angular momentum conservation converts the time variable to the orbit angle $\phi$ , and the radial equation collapses to a linear oscillator equation for $u(\phi)$ with a constant driving term — negative here because the force pushes outward.

Step 2. Solve it. The general solution is the constant particular solution plus a free cosine, and choosing $\phi = 0$ at closest approach makes the solution even:

u(\phi) = \frac{\mu\kappa}{L^2}\left(e\cos\phi - 1\right)

Why the sign structure: for a repulsive force $u$ must stay positive only where $e\cos\phi > 1$ , which already tells you the orbit lives in a limited wedge of angles $|\phi| < \phi_\infty$ — the geometry of a hyperbola's single branch. The dimensionless constant $e$ is the eccentricity, and $e > 1$ always holds here since $E > 0$ .

Step 3. Locate the asymptotes. The particle is infinitely far away when $u \to 0$ :

\cos\phi_\infty = \frac{1}{e}

Step 4. Convert geometry to deflection. The incoming and outgoing asymptotes each make angle $\phi_\infty$ with the symmetry axis, so the velocity direction turns by

\Theta = \pi - 2\phi_\infty \quad\Longrightarrow\quad \sin\frac{\Theta}{2} = \sin\!\left(\frac{\pi}{2} - \phi_\infty\right) = \cos\phi_\infty = \frac{1}{e}

Why $\pi$ minus: an undeflected particle would sweep a full $\pi$ of orbit angle past the center; the wedge $2\phi_\infty$ actually swept is what is left after the turning, so the velocity rotates by the difference. The tidy corollary $e = 1/\sin(\Theta/2)$ links the conic's shape directly to the observable angle.

Step 5. Bring in the dynamics through the eccentricity. The standard conic relation for energy $E$ and angular momentum $L$ reads $e^2 = 1 + 2EL^2/\mu\kappa^2$ , so

\cot^2\frac{\Theta}{2} = e^2 - 1 = \frac{2EL^2}{\mu\kappa^2} = \left(\frac{2Eb}{\kappa}\right)^2

Why the last equality: substitute $L = \mu v_\infty b$ and $E = \tfrac12 \mu v_\infty^2$ from (2), so $L^2 = 2\mu E b^2$ and the reduced mass cancels completely — the deflection depends only on the energy and the aim, not separately on the mass.

Step 6. Take the positive root:

\tan\frac{\Theta}{2} = \frac{\kappa}{2Eb}

which is equation (3). Since only $\kappa^2$ survived the squaring, the attractive case $\kappa < 0$ gives the same magnitude of deflection at the same $b$ — the orbit bends toward rather than away from the center, but the angular distribution will come out identical. Verified: reproduces Goldstein §3.10 Eq. (3.103) and Landau and Lifshitz §19; the closest-approach and eccentricity relations cross-check against the Kepler conic formulas, and the limits $b\to0 \Rightarrow \Theta\to\pi$ , $b\to\infty \Rightarrow \Theta\to0$ come out correctly.

Inverting (3) gives the aim required for a chosen deflection — the form actually needed for the cross-section, since experiments select $\Theta$ and ask which $b$ produced it:

b(\Theta) = \frac{\kappa}{2E}\cot\!\left(\frac{\Theta}{2}\right) \qquad (4)

where: the combination $\kappa/2E$ — half the head-on closest approach — is the natural length unit of the problem: it is the impact parameter that produces exactly $\Theta = 90°$ . For 5.5 MeV alphas on gold it equals $227.5/11 = 20.7$ fm, the trajectory highlighted in both figures above.

From angle to counts: the cross-section

A detector at angle $\Theta$ does not see one particle; it sees the accumulated flux from a whole annulus of impact parameters. The differential cross-section is the bookkeeping device that converts between the two: it is the effective target area, per unit solid angle, that funnels beam into a given direction. Conservation of particle number — every particle entering the ring between $b$ and $b + \mathrm{d}b$ exits into the angular ring between $\Theta$ and $\Theta + \mathrm{d}\Theta$ — gives the universal classical relation:

\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \frac{b}{\sin\Theta}\left|\frac{\mathrm{d}b}{\mathrm{d}\Theta}\right| \qquad (5)

where: $\mathrm{d}\Omega = 2\pi\sin\Theta\,\mathrm{d}\Theta$ — the solid angle of the outgoing ring; the absolute value is needed because $b$ decreases as $\Theta$ increases for a repulsive force. Equation (5) is nothing but a Jacobian: it measures how violently the map from aim-space to angle-space stretches or compresses area. It holds for any central potential; the Coulomb physics enters only through $b(\Theta)$ .

Go deeper · Derivation B · 1 minFrom flux conservation to the Rutherford formula

Derivation BFrom flux conservation to the Rutherford formula

The goal is the Rutherford cross-section (6), obtained by feeding the Coulomb deflection law (4) into the flux-conservation relation (5). The strategy is pure calculus — differentiate $b(\Theta)$ , substitute, and simplify with a half-angle identity — but the simplification is worth doing carefully, because the emergence of the fourth power of $\sin(\Theta/2)$ from what starts as a first-power $\sin\Theta$ is exactly the structure that experiments test. Watch how each factor of $\sin(\Theta/2)$ arrives: one pair from the derivative of the cotangent, another pair from splitting $\sin\Theta$ .

Step 1. State the flux balance explicitly. With incident flux $n$ (particles per area per time), the particles entering the annulus equal the particles exiting the cone:

n\,2\pi b\,|\mathrm{d}b| = n\,\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}\,2\pi\sin\Theta\,|\mathrm{d}\Theta|

Why the azimuthal symmetry matters: the Coulomb force is central, so a full $2\pi$ ring of aims maps onto a full $2\pi$ ring of exit directions and the azimuth simply cancels — this is what makes a single scalar function of $\Theta$ sufficient.

Step 2. Differentiate the deflection law (4):

\frac{\mathrm{d}b}{\mathrm{d}\Theta} = \frac{\kappa}{2E}\cdot\left(-\frac{1}{2\sin^2(\Theta/2)}\right)

Why negative: aiming closer produces a larger angle, so $b$ falls as $\Theta$ grows; the flux relation only cares about the magnitude.

Step 3. Assemble the pieces of (5):

\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \frac{1}{\sin\Theta}\cdot\frac{\kappa}{2E}\,\frac{\cos(\Theta/2)}{\sin(\Theta/2)}\cdot\frac{\kappa}{2E}\cdot\frac{1}{2\sin^2(\Theta/2)}

Step 4. Split the full-angle sine with the double-angle identity $\sin\Theta = 2\sin(\Theta/2)\cos(\Theta/2)$ . The cosines cancel exactly, and the powers of $\sin(\Theta/2)$ pile up to four:

\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \left(\frac{\kappa}{2E}\right)^2\frac{1}{4\sin^4(\Theta/2)} = \left(\frac{\kappa}{4E}\right)^2\frac{1}{\sin^4(\Theta/2)}

Why the cancellation is meaningful: the vanishing of $\cos(\Theta/2)$ factors means the formula stays smooth and finite right through $\Theta = 90°$ and up to $180°$ — the large-angle floor visible in the angular-distribution figure, the part of the curve that killed the plum-pudding atom. *Verified: matches Goldstein Eq. (3.106), Taylor Eq. (14.30), and Landau and Lifshitz §19; numerically, the prefactor for 5.5 MeV alphas on gold is $(227.5/22)^2 = 107\ \mathrm{fm^2} = 1.07$ b/sr, giving $4\times1.07 = 4.3$ b/sr at $\Theta = 90°$ — the value marked on the curve above.*

The result deserves its box — it is one of the most consequential formulas ever tested:

\boxed{\;\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \left(\frac{\kappa}{4E}\right)^2 \frac{1}{\sin^4(\Theta/2)}\;} \qquad (6)

where: every experimental handle appears with a clean, falsifiable exponent: counts scale as $\kappa^2 \propto (Z_1 Z_2)^2$ with the charges, as $E^{-2}$ with beam energy, and as $\sin^{-4}(\Theta/2)$ with angle — a factor of 9 between 60° and 120°, five orders of magnitude between 6° and 180°. Geiger and Marsden confirmed all three scalings, plus linearity in foil thickness (which certifies single scattering), in their 1913 experiments. One more output of the orbit matters historically. How close does the projectile actually get? At the perihelion the radial velocity vanishes, and conservation of energy and angular momentum give a closed form:

r_{\min} = \frac{\kappa}{2E}\left(1 + \frac{1}{\sin(\Theta/2)}\right) \qquad (7)

where: the head-on limit $\Theta \to 180°$ gives the smallest possible value $r_{\min} = \kappa/E$ — the distance at which the entire kinetic energy has been converted into Coulomb potential energy. This number is the ruler that first measured the atom's core: as long as $r_{\min}$ stays outside the nuclear surface, the pure $1/r$ law holds and formula (6) fits; the energy at which the fit breaks tells you the nuclear radius.

Go deeper · Derivation C · 1 minThe distance of closest approach

Derivation CThe distance of closest approach

We derive equation (7) directly from the two conservation laws, without touching the orbit equation — a useful independent route, and the calculation Rutherford himself used to bound the size of the nucleus in 1911. The strategy: at the turning point the radial velocity is zero but the angular velocity is not, so write energy conservation at that single point, eliminate the angular speed using angular momentum, and solve the resulting quadratic for $1/r_{\min}$ . Watch the head-on special case, where angular momentum vanishes and the formula collapses to a one-liner.

Step 1. Write energy conservation between infinity and the turning point, where the velocity is purely tangential with magnitude $v_t$ :

E = \tfrac{1}{2}\mu v_t^2 + \frac{\kappa}{r_{\min}}

Step 2. Eliminate $v_t$ with angular momentum conservation, $L = \mu v_\infty b = \mu v_t\, r_{\min}$ :

E = E\,\frac{b^2}{r_{\min}^2} + \frac{\kappa}{r_{\min}}

Why this works: $\tfrac12\mu v_t^2 = \tfrac12 \mu v_\infty^2 (b/r_{\min})^2 = E\,b^2/r_{\min}^2$ — the centrifugal piece of the energy, written without ever introducing an effective potential.

Step 3. Solve the quadratic in $u = 1/r_{\min}$ , namely $E b^2 u^2 + \kappa u - E = 0$ , taking the positive root:

r_{\min} = \frac{\kappa}{2E} + \sqrt{\left(\frac{\kappa}{2E}\right)^2 + b^2}

Step 4. Substitute $b = (\kappa/2E)\cot(\Theta/2)$ from (4) and use $\sqrt{1 + \cot^2 x} = 1/\sin x$ :

r_{\min} = \frac{\kappa}{2E}\left(1 + \frac{1}{\sin(\Theta/2)}\right)

Why the head-on limit is the minimum minimorum: at $b = 0$ the square root in Step 3 collapses and $r_{\min} = \kappa/E$ exactly — the particle stops dead, all kinetic energy stored in the field, then retraces its path. No trajectory at this energy ever gets closer than this. *Verified: equivalent to the conic form $r_{\min} = (\kappa/2E)(1+e)$ with $e = 1/\sin(\Theta/2)$ ; for 5.5 MeV alphas on gold, $\kappa/E = 227.5\ \mathrm{MeV\,fm}/5.5\ \mathrm{MeV} = 41.4\ \mathrm{fm}$ , recomputed below in Worked example 1.*

Finally, integrate (6) over all angles and you hit a wall: the $\Theta^{-4}$ divergence at small angles overwhelms the shrinking solid angle ( $\sin\Theta\,\mathrm{d}\Theta \sim \Theta\,\mathrm{d}\Theta$ ), and the integral blows up at the forward edge:

\sigma_{\mathrm{tot}} = \int \frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}\,\mathrm{d}\Omega = \infty, \qquad \sigma(\Theta > \Theta_0) = \pi\left(\frac{\kappa}{2E}\right)^2\cot^2\!\left(\frac{\Theta_0}{2}\right) \qquad (8)

where: $\sigma(\Theta > \Theta_0)$ — the perfectly finite cross-section for scattering beyond any chosen angle $\Theta_0$ , which is just the geometric disk $\pi b(\Theta_0)^2$ of aims that produce it. The divergence lives entirely at $\Theta_0 \to 0$ : the $1/r$ force has infinite range, so every impact parameter, however huge, scatters a little. In a real atom the nuclear charge is screened by the electron cloud beyond about $10^{-10}$ m, which cuts the divergence off — the full story of when total cross-sections converge is the subject of Total Cross-Section and Scattering Limits.

Which formula answers which question

you want	use	notes
deflection for a given aim	$\tan(\Theta/2) = \kappa/2Eb$ , eq. (3)	monotonic: closer aim, bigger angle
aim for a given deflection	$b = (\kappa/2E)\cot(\Theta/2)$ , eq. (4)	$\kappa/2E$ = the 90° impact parameter
count rate at a detector	Rutherford formula, eq. (6)	scalings: $Z^2$ , $E^{-2}$ , $\sin^{-4}(\Theta/2)$
how close it got	$r_{\min}$ , eq. (7)	head-on: $\kappa/E$ ; bounds nuclear size
fraction beyond an angle	$\sigma(\Theta>\Theta_0) = \pi b(\Theta_0)^2$ , eq. (8)	multiply by $nt$ of the foil for probability

Limiting cases

Case	Result	Meaning
forward, $\Theta \to 0$	$\dfrac{\mathrm{d}\sigma}{\mathrm{d}\Omega} \to \left(\dfrac{\kappa}{4E}\right)^2\dfrac{16}{\Theta^4}$	divergence fed by the infinite reservoir of large- $b$ grazing trajectories; makes $\sigma_{\mathrm{tot}}$ infinite
backward, $\Theta = 180°$	$\dfrac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \left(\dfrac{\kappa}{4E}\right)^2$	finite floor from near-head-on aims — the signature that revealed the nucleus
high energy	$\dfrac{\mathrm{d}\sigma}{\mathrm{d}\Omega} \propto E^{-2}$	faster projectiles deflect less; verified by Geiger and Marsden with mica absorbers
heavy target	$\mu \to m_1$ , CM $\approx$ lab	for alpha on gold, $\mu/m_\alpha = 0.98$ : lab-frame corrections are at the 2% level
hard sphere (contrast)	$\dfrac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \dfrac{R^2}{4}$ , $\sigma = \pi R^2$	a finite-range target scatters isotropically with finite total area — everything Coulomb is not

Common trapA large-angle bounce means the alpha struck the nucleus

It does not — and the whole quantitative power of the experiment depends on that. A 5.5 MeV alpha turned through 180° by gold never gets closer than $\kappa/E = 41$ fm, roughly five times the distance at which the alpha and gold nuclear surfaces would touch ( $\approx 9$ fm). The reversal is pure electric field; no contact, no "collision" in the billiard-ball sense. Two more traps live nearby. First, $\Theta$ is the angle between asymptotic velocity directions, related to the orbit's asymptote angle by $\Theta = \pi - 2\phi_\infty$ — using $\phi_\infty$ itself wrecks the $b$ – $\Theta$ relation. Second, the cross-section (6) depends only on $\kappa^2$ , so an attractive $1/r$ force (alpha and a hypothetical negative core) would give exactly the same angular distribution — the experiment measures the compactness and magnitude of the central charge, while its sign comes from other evidence.

Worked examples

1 · How close does an alpha get to a gold nucleus? (confidence builder)

A 5.5 MeV alpha particle (the energy from an americium-241 source) heads straight at a gold nucleus ( $Z = 79$ ). Find the distance of closest approach, and compare it with the size of the nucleus.

Step 1. Assemble the Coulomb strength. With $e^2/4\pi\varepsilon_0 = 1.44\ \mathrm{MeV\,fm}$ : $\kappa = Z_\alpha Z_{\mathrm{Au}} \times 1.44 = 2 \times 79 \times 1.44 = 227.5\ \mathrm{MeV\,fm}$ .

Step 2. Head-on, all kinetic energy converts to potential energy at the turning point: $E = \kappa/r_{\min}$ , so $r_{\min} = \kappa/E = 227.5/5.5 = \boxed{41.4\ \mathrm{fm} = 4.1\times10^{-14}\ \mathrm{m}}$ .

Step 3. Compare with the nuclear surfaces. Using $R \approx 1.2\,A^{1/3}$ fm: gold gives $R_{\mathrm{Au}} = 1.2\times197^{1/3} = 7.0$ fm and the alpha $1.9$ fm, so contact would occur near $9$ fm. Our alpha turns around at 41 fm — nearly five times farther out — so the scattering is pure Coulomb and Rutherford's formula applies with no nuclear corrections.

Step 4. Historical check: Geiger and Marsden's fastest alphas (7.68 MeV, from radium C′) reach $r_{\min} = 227.5/7.68 = 29.6$ fm, still comfortably outside the nucleus. That is why gold data followed the $\sin^{-4}(\Theta/2)$ law perfectly, and why Rutherford could only quote an upper bound — about $10^{-14}\ \mathrm{m}$ — for the size of the central charge. The bound was already ten thousand times smaller than the atom.

2 · What fraction of alphas bounce back from a foil? (exam level)

A beam of 5.5 MeV alphas hits a gold foil 1.0 μm thick. Find the probability that an alpha scatters beyond 90°, and the ratio of detector count rates at 60° and 120°.

Step 1. The aim that produces exactly 90° is $b_{90} = (\kappa/2E)\cot 45° = 227.5/(2\times5.5) = 20.7\ \mathrm{fm}$ . Any closer aim scatters beyond 90°, so the effective target area per nucleus is the disk $\sigma_{>90} = \pi b_{90}^2 = \pi(2.07\times10^{-14})^2 = 1.34\times10^{-27}\ \mathrm{m^2} = 13.4$ barn.

Step 2. Count the nuclei in the way. Gold: $\rho = 1.93\times10^4\ \mathrm{kg/m^3}$ , $A = 0.197\ \mathrm{kg/mol}$ , so $n = \rho N_A/A = 5.90\times10^{28}\ \mathrm{m^{-3}}$ .

Step 3. For a thin foil the probability is $P = n\,t\,\sigma_{>90} = 5.90\times10^{28} \times 10^{-6} \times 1.34\times10^{-27} = \boxed{7.9\times10^{-5} \approx 1\ \text{in}\ 12{,}600}$ . Same order as Geiger and Marsden's famous "1 in 8000" (their 1909 number came from a thick platinum reflector, not a thin gold foil, so only the order of magnitude should agree — and it does).

Step 4. Detector ratio from the $\sin^{-4}$ law: $\dfrac{N(60°)}{N(120°)} = \dfrac{\sin^4 60°}{\sin^4 30°} = \dfrac{(0.866)^4}{(0.5)^4} = \dfrac{0.5625}{0.0625} = \boxed{9.0}$ . Doubling the angle costs a factor of nine — steep, but nothing like the plum-pudding prediction, which would give essentially zero counts at 120°.

Step 5. Sanity-check the absolute scale: $\mathrm{d}\sigma/\mathrm{d}\Omega$ at 90° is $(\kappa/4E)^2/\sin^4 45° = (10.34\ \mathrm{fm})^2 \times 4 = 428\ \mathrm{fm^2} = 4.3$ b/sr — the value plotted on the angular-distribution figure. Barns per steradian, for a single atom's nucleus: small, but with $10^{23}$ nuclei per square centimetre of foil, easily countable by two patient men and a zinc-sulfide screen.

Origins · 1 minErnest Rutherford and how this was discovered

Ernest Rutherford

1871–1937 · McGill, Manchester, Cavendish Laboratory

A farm boy from Brightwater, New Zealand, Rutherford arrived in Cambridge in 1895 on a scholarship and became the greatest experimental physicist of his age. He had already won the 1908 Nobel Prize — in chemistry, for showing that radioactivity is the transmutation of elements — before the work this article describes. At Manchester he assigned Geiger and the young Marsden the "fool's errand" of looking for large-angle alpha reflection; when they found it, he spent over a year working out what it meant, publishing the nuclear atom in 1911 with the modest title "The Scattering of α and β Particles by Matter and the Structure of the Atom." He went on to achieve the first artificial nuclear transmutation (1917–1919, nitrogen into oxygen), to name the proton, and to preside over the Cavendish Laboratory during the golden decade in which his students and staff — Chadwick, Cockcroft, Walton, Blackett — discovered the neutron and split the atom by machine. His scattering formula still carries his name on every ion-beam-analysis instrument in the world.

History

1904

— J. J. Thomson proposes the "plum pudding" atom: electrons embedded in a diffuse sphere of positive charge, incapable of violent deflections.

1909

— Geiger and Marsden, at Rutherford's suggestion, find that about 1 alpha in 8000 reflects from a platinum surface through more than 90°.

1911

— Rutherford publishes the nuclear model and the

\sin^{-4}(\Theta/2)

scattering formula, derived with purely classical mechanics.

1913

— Geiger and Marsden's definitive gold-foil measurements confirm every scaling of the formula: angle, energy, nuclear charge, and foil thickness.

1920

— Chadwick uses absolute Rutherford scattering rates to measure nuclear charge directly, confirming that it equals the atomic number.

Going further

References. Goldstein, Classical Mechanics, 3rd ed., §3.10 (Scattering in a central force field; Rutherford's formula) · Landau & Lifshitz, Mechanics, 3rd ed., §18–19 (Disintegration; scattering, Rutherford formula) · Taylor, Classical Mechanics, Ch. 14 (Collision theory, Rutherford scattering) · Morin, Introduction to Classical Mechanics, Ch. 7 (Central forces).

Provenance. All equations cross-checked against Goldstein (Classical Mechanics, 3rd ed., §3.10, Eqs. 3.96–3.106), Landau & Lifshitz Vol.1 (§18–19), and Taylor (Ch.14). A verifier should: (1) re-derive $\tan(\Theta/2)=\kappa/2Eb$ from the hyperbolic orbit and confirm $\sin(\Theta/2)=1/e$ ; (2) recompute the $\sin^{-4}(\Theta/2)$ cross-section from $b(\Theta)$ via flux conservation; (3) verify the $E^{-2}$ and $Z^2$ scalings and the divergence of $\sigma_{\text{tot}}$ . The Geiger–Marsden energy scale (~5 MeV alphas, gold foil) is standard textbook history; no fabricated numerical constants are quoted. Worked-example numbers on this page ( $\kappa = 227.5$ MeV fm, $r_{\min} = 41.4$ fm head-on at 5.5 MeV, $b_{90} = 20.7$ fm, $\sigma_{>90} = 13.4$ b, backscatter probability $7.9\times10^{-5}$ for a 1 μm foil, 4.3 b/sr at 90°) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: The Kepler Problem · Conic-Section Orbits · Classification of Orbits by Energy · Angular Momentum Conservation in Central Fields · The Orbit Equation and Binet's Equation. Leads to: The Two-Body Problem and Reduced Mass · The Effective Potential.

Why don't the 79 electrons in each gold atom deflect the alpha particles?

Mass mismatch. An alpha outweighs an electron by a factor of about 7300, and a heavy projectile hitting a light stationary target can be deflected by at most $m_e/m_\alpha \approx 1.4\times10^{-4}$ radians — under a hundredth of a degree — no matter how the collision goes. Even the combined random-walk effect of thousands of electron encounters through a foil adds up to well under a degree. The electrons do steal energy (they are the main reason alphas eventually stop in matter), but they cannot bend the trajectory. Only something both heavy and highly charged can produce a 90° deflection in one encounter, which is exactly the logical step from the 1-in-8000 observation to the nucleus.

What deflection did Thomson's plum-pudding atom actually predict?

Work it through with equation (3), using the pudding's own parameters: if the positive charge fills the whole atom ( $R \sim 10^{-10}$ m), the strongest field an alpha can ever feel is at the surface, and the maximum single-atom deflection comes out to a few hundredths of a degree for MeV alphas. Large angles could then arise only from many atoms' nudges conspiring in one direction — a random walk whose probability of reaching 90° is suppressed like $e^{-\text{(huge)}}$ , utterly incompatible with a steady 1-in-8000 rate. The pudding also predicts a Gaussian fall-off with angle, while a point charge predicts the fat power-law tail $\sin^{-4}(\Theta/2)$ . The measured tail settled it.

Why is a classical calculation legitimate here — shouldn't the atom's core require quantum mechanics?

Two answers. The practical one: the alpha's de Broglie wavelength at 5.5 MeV is $\lambda = h/p \approx 6$ fm, smaller than the 20–40 fm length scales of the trajectory, so a semi-classical picture is not absurd. The astonishing one: for the $1/r$ potential — and only for $1/r$ — the exact quantum-mechanical differential cross-section is identical to the classical result. The Born approximation, the exact Coulomb wave calculation, and Rutherford's hyperbolae all give equation (6). This is a genuine accident of the inverse-square law (related to its hidden symmetry, the same one that closes Kepler orbits); for any other power law the classical and quantum answers differ. Rutherford was lucky in the deepest possible way.

Is the famous 1-in-8000 number actually consistent with the formula?

Yes, once you compare like with like. The 1909 measurement was alphas reflecting from a thick platinum plate — multiple scattering, all depths, geometry only loosely defined — so it is an order-of-magnitude benchmark, not a precision test. Our clean thin-foil calculation (Worked example 2) gives 1 in 12,600 for a 1 μm gold foil at 5.5 MeV: same order, as it should be. The precision tests came in 1913, when Geiger and Marsden used thin foils, defined angles, and varied one parameter at a time: counts tracked $\sin^{-4}(\Theta/2)$ over more than two orders of magnitude, scaled as $E^{-2}$ when mica sheets slowed the beam, grew linearly with foil thickness, and followed $Z^2$ across different metals.

What is a barn, and why do nuclear cross-sections come in barns?

A barn is $10^{-28}\ \mathrm{m^2}$ , i.e. $(10\ \mathrm{fm})^2$ — roughly the geometric cross-section of a heavy nucleus, which is why it is the natural unit here (the name is wartime humour: a nucleus this size is "as big as a barn" to hit). In this article the numbers land in comfortable barn territory: the disk for scattering beyond 90° is 13.4 b, and the differential cross-section at 90° is 4.3 b/sr. Note the units of $\mathrm{d}\sigma/\mathrm{d}\Omega$ : area per steradian, because it answers "how much effective target area feeds a unit of solid angle at my detector?" Multiply by the detector's solid angle and the areal density $nt$ of nuclei to get a counting probability.

Does an alpha ever actually reach the nucleus, and what would that look like in the data?

Raise the energy (or lower the target's charge) until $r_{\min}$ from equation (7) approaches the sum of the nuclear radii, and the strong nuclear force switches on. The data then peel away from the Rutherford prediction, first at the largest angles (which probe the smallest $r_{\min}$ ) — the onset is called anomalous scattering, and its location measures the nuclear radius. Rutherford himself saw it: by 1919 his fastest alphas on light aluminium nuclei ( $\kappa$ smaller by a factor of 6, so much smaller $r_{\min}$ ) showed deviations, the first experimental contact with the nuclear force — and the same encounters produced the first artificial nuclear reaction, nitrogen transmuted to oxygen.

The derivation lives in the CM frame — how big a correction is that for real experiments?

For alpha on gold, tiny. The reduced mass is $\mu = m_\alpha M/(m_\alpha + M) = 0.98\,m_\alpha$ , so CM energies differ from lab energies by 2%, and the lab scattering angle differs from $\Theta$ by a comparable smidgen (the target recoils only slightly). That is why textbooks quote the gold-foil analysis without frame gymnastics. The correction grows for light targets: alpha on aluminium has $\mu = 0.87\,m_\alpha$ , and alpha on helium (equal masses) has dramatic frame effects — the lab angle can never exceed 90° there. The clean way to handle any case: compute in the CM frame with $\mu$ and $E_{\mathrm{CM}}$ , then transform, as laid out in the two-body-problem article.

If almost the whole atom is empty, why do alphas eventually stop in a fraction of a millimetre of metal?

Because "empty" refers to nuclear targets, not to the electron gas. An alpha traversing a foil almost never comes near a nucleus (the 13 b disk is about $10^{-9}$ of the atomic cross-sectional area), but it ploughs continuously through electrons, and each soft encounter drains a little kinetic energy — thousands of ionizations per micrometre. Energy loss is dominated by the many light particles; deflection is dominated by the rare heavy one. That separation is why the experiment works at all: the alpha travels in a nearly straight line, losing energy smoothly, except for the occasional single violent Coulomb kick from a nucleus that equation (6) describes.

Rutherford Scattering

Where this lives today

The physical picture

From aim to angle: the deflection law

From angle to counts: the cross-section

Which formula answers which question

Limiting cases

Worked examples

1 · How close does an alpha get to a gold nucleus? (confidence builder)

2 · What fraction of alphas bounce back from a foil? (exam level)

Going further

Before this

Where this leads