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Impact parameter and scattering angle, $d\sigma/d\Omega = (b/\sin\theta)|db/d\theta|$, hard-sphere and Rutherford scattering, total cross sections and counting rates.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to relate impact parameter to scattering angle, compute differential and total cross sections, and predict counting rates.
You know unbound orbits are hyperbolas with the force center at a focus, and that energy and angular momentum fix an orbit. You can compute solid angles, $d\Omega = \sin\theta\,d\theta\,d\phi$. This lesson turns orbits inside out: instead of following one body, it asks where a whole beam of them goes when it hits a target, which is how physicists have learned what atoms, nuclei and quarks look like.
| Term | What it means |
|---|---|
| Impact parameter | $b$, the perpendicular distance between an incoming particle's line and the target's center. |
| Scattering angle | $\theta$, the angle between the incoming and outgoing directions. |
| Solid angle | $d\Omega = \sin\theta\,d\theta\,d\phi$, measured in steradians; the whole sphere is $4\pi$. |
| Differential cross section | $d\sigma/d\Omega$, the beam area sent into each unit of solid angle. |
| Total cross section | $\sigma$, the whole area of beam that is scattered at all. |
| Barn | $10^{-28}$ m², the unit of nuclear cross sections. |
| Luminosity | Beam rate times target density: the rate is luminosity times cross section. |
Fire a broad parallel beam at a fixed target. Each particle arrives with some impact parameter $b$, the distance by which its line would miss the center, and leaves at a scattering angle $\theta$ that depends on $b$. For a central force the relation $b(\theta)$ is the whole physics.
The particles arriving in a thin ring of radius $b$ and width $db$, area $2\pi b\,db$, leave into a cone between $\theta$ and $\theta + d\theta$, of solid angle $2\pi\sin\theta\,d\theta$. The ratio is the differential cross section,
$$\frac{d\sigma}{d\Omega} = \frac{b}{\sin\theta}\left|\frac{db}{d\theta}\right|,$$
an area per steradian. A detector covering solid angle $\Delta\Omega$ at angle $\theta$ counts
$$N = I\,n_A\,\frac{d\sigma}{d\Omega}\Delta\Omega,$$
with $I$ the beam rate and $n_A$ the target atoms per unit area. Integrating over all directions gives the total cross section $\sigma$, the effective area the target presents.
Another way: picture
Think of rain falling on a round umbrella. The drops that hit are those within the umbrella's outline, an area $\pi R^2$ however the umbrella is curved. Where they splash depends on where they hit: near the rim they glance off, near the center they bounce back up. Counting splashes in each direction measures the umbrella's shape without ever seeing it.
Another way: steps
A point particle bouncing off a fixed hard sphere of radius $R$ hits it at an angle of incidence $\alpha$ with $b = R\sin\alpha$, and reflects symmetrically, leaving at $\theta = \pi - 2\alpha$. So $b = R\cos(\theta/2)$. Then $|db/d\theta| = \tfrac{R}{2}\sin(\theta/2)$, and using $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$,
$$\frac{d\sigma}{d\Omega} = \frac{R\cos(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} \cdot \frac{R}{2}\sin\frac{\theta}{2} = \frac{R^2}{4}.$$
Hard spheres scatter isotropically, equally in every direction, which is a surprise: one might expect more backscattering. Integrating over $4\pi$ gives $\sigma = \pi R^2$, the area of the sphere's shadow. For two spheres of radii $R_1$ and $R_2$, the relevant radius is $R_1 + R_2$, which is how the sizes of gas molecules were first estimated from how often they collide.
For a repulsive Coulomb force $F = k/r^2$, the orbit is a hyperbola, and working out its asymptotes gives $b = \frac{k}{2E}\cot(\theta/2)$, with $E$ the incoming kinetic energy. Differentiating,
$$\frac{d\sigma}{d\Omega} = \left(\frac{k}{4E}\right)^2\frac{1}{\sin^4(\theta/2)}.$$
The cross section falls steeply with angle: at $60°$ it is sixteen times its value at $180°$. The quantity $k/E$ is the distance of closest approach in a head-on collision, where the particle stops and turns back. In 1909 Hans Geiger and Ernest Marsden found alpha particles bouncing back from gold foil at rates that fitted this formula, and Rutherford concluded in 1911 that an atom's positive charge sits in a tiny nucleus.
Integrate Rutherford's formula over all angles and the result diverges at $\theta \to 0$: the total cross section is infinite. The reason is physical. The Coulomb force has no edge, so a particle passing at any distance, however large, is deflected slightly. Every particle in an infinitely wide beam is scattered by some tiny angle.
In real atoms the nucleus's charge is screened by the electrons beyond about $10^{-10}$ m, which cuts off the small-angle divergence. In practice experiments only count particles scattered beyond some minimum angle, and the cross section for that is finite. The same infinite range makes long-range forces tricky throughout physics, from plasma physics to the scattering of charged particles in accelerators.
A thin target with $n_A$ atoms per unit area blocks a fraction $n_A\sigma$ of the beam, as long as that fraction is small so the atoms do not shadow one another. The reaction rate is $R = I\,n_A\,\sigma$. Particle physicists combine the beam and target factors into the luminosity $\mathcal{L}$, so that $R = \mathcal{L}\sigma$.
Nuclear cross sections are measured in barns, $10^{-28}$ m², roughly the cross-sectional area of a uranium nucleus; the name came from physicists at Purdue during World War II, joking that a uranium nucleus was as big as a barn to a neutron. Cross sections in particle physics are femtobarns and smaller, which is why colliders need enormous luminosities to see rare events.
The formulas above assume the target does not move. When the target recoils, as a proton does when hit by a neutron, the calculation is done in the center-of-mass frame, where the two-body problem reduces to one body of reduced mass, exactly as in the two-body lesson. The scattering angle there, $\theta_{\text{cm}}$, is related to the laboratory angle by the relative velocity of the frames.
For equal masses the relation is simple: $\theta_{\text{lab}} = \theta_{\text{cm}}/2$, so particles never scatter backward in the laboratory, and after an elastic collision the two leave at right angles. That is what pool players see on every clean hit, and it is why neutron moderators in reactors use hydrogen-rich water: equal masses take the most energy from the neutron in each collision.
Checking an answer. A differential cross section must have units of area. It must be positive at every angle. For a hard sphere the total must be the geometric shadow. For Rutherford scattering the backscattered rate must rise as the energy falls, as $1/E^2$, since slow particles are turned more easily. And a thin-target rate must be a small fraction of the beam.
Almost everything known about the structure of matter at small scales came from scattering experiments. Rutherford's alphas found the nucleus. In the late 1960s, electrons from the two-mile linear accelerator at SLAC in California scattered off protons at large angles far more often than a smooth proton would allow, the signature of point-like constituents inside: the quarks. Neutron scattering at Oak Ridge in Tennessee maps the arrangement of atoms in new materials.
Each experiment follows the pattern of this lesson: measure how many particles go into each direction, compare with $d\sigma/d\Omega$ calculated from a model of the force, and adjust the model until the two agree. Where they disagree, as they did for Rutherford, there is something new to find.
It would seem simpler to count particles per unit of scattering angle $\theta$. The trouble is that the ring of directions between $\theta$ and $\theta + d\theta$ is not the same size at every angle: near $\theta = 90°$ it is a wide band around the sphere, and near $0°$ or $180°$ it shrinks to a tiny cap. A detector of fixed size sees a fixed solid angle, $\Delta\Omega = A/r^2$ for a detector of area $A$ at distance $r$, wherever it is placed. Counting per steradian therefore compares like with like.
This is also why the factor $\sin\theta$ appears in the denominator of $d\sigma/d\Omega$. The hard sphere shows its effect: the rings of incoming beam near the edge, at large $b$, are large, and they send particles into the forward directions; the rings near the center are small, and they send particles backward. The two effects exactly cancel the changing size of the rings of outgoing directions, which is why the hard sphere's scattering comes out the same in every direction. When you set up a scattering calculation, always pair a ring of impact parameters with its ring of solid angle, and the factors take care of themselves.
Materials scientists use Rutherford scattering to measure thin films without cutting them. A beam of helium ions at about $2$ MeV strikes a sample, and a detector at a large angle, often near $170°$, records the ions that bounce back. Heavy atoms return the ions with nearly all their energy, light atoms with less, and atoms deeper in the film return them after losing energy on the way in and out.
The count rate at each energy follows $d\sigma/d\Omega$ from this lesson, which scales as the square of the target atom's charge, so the spectrum gives the composition and thickness of each layer. National laboratories such as Sandia in New Mexico and many university accelerator labs run this analysis for the semiconductor industry, measuring films a few nanometers thick; the method works because Rutherford's formula is exact for these energies.
A nuclear reactor runs on neutron cross sections. Uranium-235 has a fission cross section of about $580$ barns for slow neutrons but only about $1$ barn for the fast ones fission produces, so reactors slow the neutrons down in a moderator. Water works because hydrogen's mass nearly matches the neutron's, so each collision removes a large share of the neutron's energy.
Control rods contain boron or cadmium, whose absorption cross sections for slow neutrons are thousands of barns, so pushing them in soaks up neutrons and slows the chain reaction. The rate of each process in the core is flux times density times cross section, the formula of this lesson. The cross sections themselves are measured at facilities such as the Los Alamos Neutron Science Center and compiled by the National Nuclear Data Center at Brookhaven in New York.
The total cross section of a hard sphere is the area it presents to the beam, $\pi R^2$, not its surface area $4\pi R^2$. A particle either passes within $R$ of the center or it does not; the curvature of the surface affects only where it goes afterwards.
A second error is to think a cross section is always the size of the target. For a long-range force it can be much larger, as the Coulomb cross section shows; for a neutron and a nucleus it can depend strongly on the energy, rising sharply at resonances. It is an effective area for a particular process, not a physical size.
Write the impact parameter of a hard sphere.
$b = R\cos\dfrac{\theta}{2}$
From reflection at the surface.
Differentiate with respect to the angle.
$\left|\dfrac{db}{d\theta}\right| = \dfrac{R}{2}\sin\dfrac{\theta}{2}$
Chain rule.
Form the differential cross section.
$\dfrac{d\sigma}{d\Omega} = \dfrac{R\cos(\theta/2)}{\sin\theta} \cdot \dfrac{R}{2}\sin\dfrac{\theta}{2}$
$(b/\sin\theta)|db/d\theta|$.
Simplify with the double-angle formula.
$\dfrac{d\sigma}{d\Omega} = \dfrac{R^2}{4}$
$\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$.
Integrate over the sphere of directions.
$\sigma = \dfrac{R^2}{4} \times 4\pi = \pi R^2$
The geometric shadow.
A beam of $6.0 \times 10^{12}$ protons per second hits a foil with $5.0 \times 10^{22}$ atoms/m². The cross section is $2.0$ barns. Write the rate formula.
$R = In_A\sigma$
Thin target.
Convert the cross section.
$\sigma = 2.0 \times 10^{-28}\ \text{m}^2$
A barn is $10^{-28}$ m².
Multiply the numbers.
$6.0 \times 5.0 \times 2.0 = 60$
Separate from the powers of ten.
Combine the powers of ten.
$10^{12 + 22 - 28} = 10^{6}$
Add the exponents.
State the rate.
$R = 6.0 \times 10^{7}\ \text{s}^{-1}$
Sixty million reactions per second.
Alphas of $E = 5.0$ MeV meet gold, with $k = 2 \times 79 \times 1.44 = 228$ MeV fm. Find the head-on closest approach.
$d = \dfrac{k}{E} = \dfrac{228}{5.0} = 45.5\ \text{fm}$
All the kinetic energy becomes Coulomb energy.
Find the impact parameter for $90°$ scattering.
$b = \dfrac{d}{2}\cot 45° = 22.8\ \text{fm}$
$b = (k/2E)\cot(\theta/2)$.
Find the cross section for scattering beyond $90°$.
$\sigma = \pi b^2 = \pi \times 22.8^2 = 1630\ \text{fm}^2$
All particles with smaller $b$.
Convert to barns.
$1630\ \text{fm}^2 = 1630 \times 10^{-30}\ \text{m}^2 = 16.3\ \text{b}$
One fm² is $10^{-30}$ m².
Find the fraction scattered beyond $90°$ by a foil with $6 \times 10^{22}$ atoms/m².
$n_A\sigma = 6 \times 10^{22} \times 1.63 \times 10^{-27} = 1.0 \times 10^{-4}$
About one alpha in ten thousand bounces back.
Compare with the nucleus.
$d = 45.5\ \text{fm} > R_{\text{Au}} \approx 7\ \text{fm}$
The alphas never touch the nucleus, so the pure Coulomb law holds.
Write the differential cross section.
$\dfrac{d\sigma}{d\Omega} = \dfrac{R^2}{4} = \dfrac{9.0}{4}$
Isotropic scattering.
Evaluate the differential cross section.
$\dfrac{d\sigma}{d\Omega} = 2.25\ \text{mm}^2/\text{sr}$
The same at every angle.
Find the total cross section.
A parallel beam of tiny pellets is fired at a fixed hard sphere of radius $2$ mm. What is the total cross section for a pellet to be scattered?
Complete the worked solution: alpha particles of $4$ MeV approach gold nuclei, for which $k = q_1q_2/4\pi\varepsilon_0 = 200$ MeV fm. Find the head-on distance of closest approach $d$ in fm, the impact parameter for scattering through $90°$ in fm, and the cross section for scattering beyond $90°$ as a multiple of $\pi$ fm².
Set the kinetic energy equal to the Coulomb energy at closest approach.
$E = \dfrac{k}{d} \Rightarrow d =$ d
Head on, the particle stops momentarily and turns back.
Use the Rutherford relation at ninety degrees.
$b = \dfrac{d}{2}\cot 45° =$ b
$\cot 45° = 1$.
Square that impact parameter for the disk's area.
$\sigma(\theta > 90°) = \pi b^2 = \pi \times$ a
Every particle aimed closer than $b$ scatters further.
Check that the scale is nuclear.
$\text{closest approach} \sim \text{tens of femtometers}$
Larger than a gold nucleus, about $7$ fm, so the pure Coulomb law holds.
Match each scattering quantity to its description.
| offset of the incoming line from the center | $(b/\sin\theta)|db/d\theta|$ | the integral over all solid angle | proportional to $1/\sin^4(\theta/2)$ | |
|---|---|---|---|---|
| impact parameter | ||||
| differential cross section | ||||
| total cross section | ||||
| Rutherford cross section |
In a Rutherford experiment, the differential cross section at $\theta = 180°$ is $5$ (in some unit). Fill in its value at $180°$, $90°$ and $60°$ in the same unit.
| $d\sigma/d\Omega$ | |
|---|---|
| at 180° | |
| at 90° | |
| at 60° |
A point particle bounces elastically off a fixed hard sphere of radius $5$ cm. Write its impact parameter $b$, in cm, as a formula in the scattering angle $t$ (in radians).
Answer:
A beam of $4 \times 10^{12}$ particles per second strikes a thin foil holding $5 \times 10^{22}$ target atoms per square meter. The cross section for a particular reaction is $3 \times 10^{-28}$ m² (that is, $3$ barns). How many reactions occur per second, in millions?
Answer: million reactions per second
An ion-beam analysis lab at a national laboratory in New Mexico aims helium ions at a thin gold film. A detector at $60°$ counts $9000$ scattered ions per minute. If the same detector is moved to $120°$, how many per minute should it count?
Answer: counts per minute
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A point particle bounces elastically off a fixed hard sphere of radius $5$ cm. Write its impact parameter $b$, in cm, as a formula in the scattering angle $t$ (in radians).
Answer:
You can analyze scattering with cross sections. Explain to someone how counting bounced alpha particles showed that atoms have a tiny nucleus.
19. Your turn: a hard sphere has radius $3.0$ mm. Find its differential and total cross sections., step 3
$\sigma = \pi R^2 = 9.0\pi = 28.3\ \text{mm}^2$
The shadow of the sphere.