Rutherford found his formula in 1911 from classical orbits, and quantum mechanics, remarkably, gives exactly the same cross section. The quickest way there is the Born approximation for the screened potential $\beta e^{-\mu r}/r$ with $\beta=Z_{1}Z_{2}e^{2}/4\pi\varepsilon_{0}$, letting the screening length $1/\mu$ grow without limit: $d\sigma/d\Omega=(\beta/4E)^{2}/\sin^{4}(\theta/2)$. The screening is not only a device. In real matter the atomic electrons screen the nucleus, and that removes the divergence of the bare formula at small angles. The pure Coulomb field is special twice over: it falls so slowly that the total cross section is infinite, and its waves never become free, carrying a phase that grows like the logarithm of the distance. Solved exactly, it still gives Rutherford’s formula (stated here) — a case where the Born approximation happens to be exact for the cross section.

A comet swinging past the Sun follows a hyperbola whose bend depends on how close it comes; comets arriving at random would spread out in exactly Rutherford’s pattern. Quantum mechanics, rather surprisingly, leaves that pattern unchanged.

The Rutherford cross section against the scattering angle on a logarithmic scale — the base-10 logarithm of $d\sigma/d\Omega$ relative to its value straight back — for a bare Coulomb field, and for one screened at the distance $1/\mu=5/k$. The bare one grows without limit as $\theta\to0$; screening caps it.

Two charges $Z_{1}e$ and $Z_{2}e$, with $\beta =Z_1{Z}_2{e}^2\mathrm{/}4$, at the energy , scatter with the Rutherford cross section

• It is the same classically, in the Born approximation, and exactly (stated); it does not depend on the sign of $\beta$.
• It grows without limit as $\theta\to0$, and the total cross section is infinite: the field reaches arbitrarily far.
• Screening by atomic electrons, $\beta e^{-\mu r}/r$, caps it: the forward cross section becomes finite, .

Take the Born amplitude of the screened potential and let the screening length grow without limit. What is left depends on the angle only through $q^{2}=4k^{2}\sin^{2}(\theta/2)$; squaring and writing gives the formula, from which has cancelled.

The Born amplitude of $\beta e^{-\mu r}/r$ from the last lesson.

$\hbar^{2}k^{2}=2mE$.

No $\hbar$ is left: the classical result.

Near $\theta=0$ the integrand behaves as $16(\beta/4E)^{2}/\theta^{3}$.

The Coulomb cross section falls as the fourth power of $\sin(\theta/2)$, and as the square of the energy.

Rutherford’s formula contains no $\hbar$: classical orbits, the Born approximation and the exact quantum solution all give it.

The cross section goes as $1/E^{2}$: faster particles are deflected less.

The combination $\beta/4E$ is a length, a quarter of the closest approach in a head-on collision; it sets the size of the cross section.

Great work! You now know how to:

• derive Rutherford’s formula as a limit of the screened potential
• explain why the Coulomb cross section is infinite
• say how screening and the nuclear force change it