Rutherford scattering

Rutherford scattering is an elastic particle scattering process where an alpha particle scatters off of an atomic nucleus. It is an electromagnetic scattering process with both classical and quantum formulations. The process is

where $\ce{X}$ represents the nucleus.

Kinematics#

We assume that the scattering happens in relativistic regime, so we use four-vectors. The four-momentum is

The alpha particle collides with the nucleus elastically, so momentum is conserved. If we use a laboratory frame of reference in which the nucleus is at rest, the initial four-momenta are

for the alpha particle and nucleus, respectively. Then, due to conservation:

where we used a prime to denote the state after the scattering. Then

Same goes for the primed momenta. The square norms are related to the rest masses of the particles by

Since the scattering is elastic, the particles don't change and so these also don't change, $\lvert p_{\alpha} \rvert^{2}=\lvert p_{\alpha}' \rvert^{2}$ and $\lvert p_{N} \rvert^{2}=\lvert p_{N}' \rvert^{2}$. This allows us to state

Now, we can assume that $p_{\alpha}$ and $p_{N}$ are known. $p_{\alpha}'$ can also be measured with a detector. $p_{N}'$ isn't known, but we can use $(1)$ to express it in terms of all the others:

If we place ourselves in the lab frame, where the nucleus is at rest before the collision, we can write the resulting momenta as

We've chosen to write $p_{N}'$ in terms of the other three, so we're not going to need it. Knowing this, if we multiply both sides of $(2)$ by $c^{2}$ we get

If we assume that the energy of the total energies $E_{\alpha}$ and $E_{\alpha}'$ are much higher than the rest energy $m_{\alpha}c^{2}$, then we can ignore the rest energy. The left hand side becomes $(p_{\alpha}\cdot p_{N})c^{2}=E_{\alpha}Mc^{2}$. In the right hand side, we can write the scalar product explicitly as $\mathbf{p}'_{\alpha}\cdot \mathbf{p}_{\alpha}=p'_{\alpha}p_{\alpha}\cos \theta$, where $\theta$ is the angle between the momentum vectors. In the high-energy regime we're in, the energy is approximately $E=\sqrt{ p^{2}c^{2}+m^{2}c^{4} }\simeq \sqrt{ p^{2}c^{2} }=pc$, so $p_{\alpha}'p_{\alpha}\cos \theta \simeq(E_{\alpha}'E_{\alpha}/c^{2})\cos \theta$. Put this in the right hand side to find

Extract the alpha particle's energy to get

where $\theta$ is the scattering angle. This quantity can be measured through a detector, so it can be useful to invert the relation to find a different quantity in terms of this one.

Cross section#

Rutherford scattering involves a particle coming close to a nucleus, then getting pushed or pulled by electromagnetic forces. This of course occurs whenever the trajectory of the particle comes close enough to the nucleus for the repulsion to be significant. When aiming a particle beam at a target, most of the time the incident particles will be too far away from the nucleus to be significantly affected by it. To express the Probability of a particle being scattered by the nucleus, we define the Rutherford cross section. There's two paths to reach this value, a classical one a quantum one. Both are presented below.

Classical method#

We'll use a generic particle as the projectile. Historically, Rutherford used alpha particles in his first experiments on the subject and today electrons are especially used in real experiments because, being elementary particles, they don't have an internal structure, which makes some subsequent conclusions easier. This said, the results are largely only dependent on the charge of the projectile, as long as we assume it's smaller then the nucleus by a good margin. The only major difference is that negatively-charged particles are attracted by the nucleus, not repelled, so the scattering angle flips sign and goes in the opposite direction compared to positively-charged ones. The magnitude, however, stays largely the same.

Say our projectile has charge $ze$ and the nucleus has charge $Ze$, where $e$ is the elementary charge. We shoot the projectile towards the nucleus with initial velocity $\mathbf{v}_{0}$.

Our trajectory isn't likely to be head-on. The distance between the trajectory before scattering and the trajectory that would lead to a head-on collision¹ is known as the impact parameter $b$. Once the projectile gets close to the nucleus, it'll be repelled or attracted by Coulomb's law². As in all non-bound trajectories drawn by a force that goes like $1/r^{2}$, it ends up being hyperbolic.

To predict the trajectory, we use energy arguments. When far from the nucleus, the projectile experiences no electrostatic force and is a typical free particle. The potential energy is $U=0$ and the kinetic energy is $T=\frac{1}{2}mv_{0}^{2}$. The angular momentum with respect to the center of the nucleus is $L=|\mathbf{r}\times m\mathbf{v}|=mv_{0}b$.

When the projectile approaches the nucleus, it starts to feel the electrostatic force. It'll reach a point of closest approach, at a distance $r_\text{min}$, and then distance itself. $r_\text{min}$ is dependent on $b$; when $b=0$, you have a head-on collision, which leads to the lowest possible distance $d$. Which can calculate this by using conservation of energy and noting that in $d$, the kinetic energy must be zero (it's inverting direction) and all the energy left is electrostatic potential energy. So

You can extract $d$ by inverting this:

In most cases, energy conservation will gives us a mixture of kinetic and potential energy:

This interaction inherently has cylindrical symmetry. This is because the only geometrical property that the scattering angle depends on is the impact parameter. This means that it behaves the same regardless of where the projectile's trajectory is around the nucleus.

In fact, all particles incident on some ring $[b,b+db]$ around the nucleus are scattered into an angular region $d\theta$. This ring has area $2\pi bdb$. A real target has a large number of nuclei and each of them has this ring around it. Call $df$ be the fraction of all incident particles that are going through this ring around some nucleus. To determine it, we need the number of nuclei, specifically their density over the impact surface. If $n$ is the volume density of nuclei and $x$ is the thickness of the target (you can imagine it as a thin foil), the surface density of nuclei will be $nx$. Thus, the fraction will be

The fraction of all particles passing within a ring of radius $b$ (area $\pi b^{2}$) is instead

If we consider the nucleus to be heavy enough with respect to the projectile to ignore recoil, then the initial and final momenta of the projectile, $\mathbf{p}_{i}$ and $\mathbf{p}_{f}$, are the same in magnitude. The only difference is the trajectory. Far from the nucleus, the magnitude is $\lvert \mathbf{p}_{i} \rvert=mv_{0}=\lvert \mathbf{p}_{f} \rvert$.

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We use instantaneous coordinates $\mathbf{r}$ and $\beta$, where $\beta$ is the angle between the bisector and $\mathbf{r}$, with $\mathbf{r}$ locating the particle. Then

assuming $|\mathbf{p}_{i}|=|\mathbf{p}_{f}|$. Now apply Newton’s second law $F=dp/dt$:

At $t=0$, $\beta=- (\pi-\theta)/2$; at $t=\infty$, $\beta= (\pi-\theta)/2$.

Compute instantaneous velocity $\mathbf{v}$:

Only the tangential component contributes near the nucleus:

By conservation of angular momentum:

Substitute into $(4)$:

Combine with $(3)$:

Then combine with $(2)$:

Now find the cross section:

where $nx$ is the number of nuclei per unit area and $df/d\Omega$ is the number of incident particles per solid angle. For annular geometry, $d\Omega=2\pi\sin\theta d\theta$.

Using $(1)$ and $(5)$:

Thus, the Rutherford cross section is:

It scales as $\propto z^{2}T^{-2}\sin^{-4}(\theta/2)$. These dependencies were experimentally confirmed by Geiger and Marsden.

Quantum Method#

In the quantum regime, the uncertainty principle applies. Uncertainty $\Delta b$ in $b$ implies $\Delta p\sim \hbar/\Delta b$. The classical derivation holds if $\Delta b\ll b$ and $\Delta p\ll p$, i.e.

If $b\sim0$ and projectile energy is very high, the strong force becomes relevant, and Rutherford scattering breaks down. This can be used to estimate the nuclear radius.

To compute the Rutherford cross section via quantum mechanics, assume:

• nuclear recoil is negligible.
• initially, a point charge impinges on the atomic nucleus.
• the Born approximation: $Z\alpha\ll1$, where $\alpha=1/137$ is the Fine-structure constant.

The incoming and outgoing electron wave functions are plane waves:

with $V$ the normalization volume. Electrons are free at $t=0$ and $t=+\infty$.

Treat discrete electron structures as continuous:

where $N$ is the number of scattering centers (effectively nucleons) and $n$ is the density. For large volume:

Fermi’s golden rule gives the reaction rate:

Final-state energy $E_{f}$ satisfies $dE_{f}=dE'$.

The phase space density is:

The cross section for the electron into solid angle $d\Omega$ is:

Substitute $v \to c$ (relativistic limit, $|\mathbf{p'}|\sim E'/c$):

The interaction Hamiltonian is:

where $\phi(\mathbf{r})$ is the electric potential. The transition matrix is:

Define the transferred momentum $\mathbf{q}\equiv\mathbf{p}-\mathbf{p}'$. Then:

and

since $|\mathbf{p'}|=|\mathbf{p}|$ and $1-\cos\theta=2\sin^{2}(\theta/2)$. Then:

For a point charge with spherical symmetry:

The matrix becomes:

Switch to spherical coordinates ($\alpha$: polar, $\phi$: azimuthal):

with $x=qr/\hbar$. This diverges. Introduce $e^{-\lambda x}$:

As $\lambda \to 0$, this gives 1. Then:

The matrix becomes:

The integral is the Form factor $F(\mathbf{q})$.

For a point charge:

and the cross section is:

In the relativistic limit ($E'=|\mathbf{p'}|c$):

In the non-relativistic limit ($p=mv$, $E_{cin}=\frac{1}{2}mv^{2}$, $E'\sim mc^{2}$):

matching the classical result.