1.1 Rutherford scattering and the nuclear atom

(a) The experiment and the nuclear atom

In the Geiger–Marsden experiment a collimated beam of \(\alpha\) particles (He\(^{2+}\) nuclei of a few MeV) from a radioactive source strikes a gold foil about \(10^{-7}\)–\(10^{-6}\) m thick, and the scattered particles are counted by the scintillations they produce on a zinc sulphide screen at angle \(\theta\) (Figure 1.1). Most particles pass undeflected, a few are deflected appreciably and about one in 20 000 is turned back through more than \(90^\circ\). A diffuse Thomson atom cannot do this, so the positive charge and nearly all the mass of the atom must be concentrated in a nucleus of radius \(10^{-15}\)–\(10^{-14}\) m, against an atomic radius of about \(10^{-10}\) m, so that \(R_{\text{atom}}/R_{\text{nucleus}}\sim10^4\)–\(10^5\).

The analysis assumes point charges interacting only by the Coulomb force, a fixed (infinitely heavy) nucleus, a single collision per particle with negligible effect of the electrons, and non-relativistic classical motion. A projectile of charge \(ze\), mass \(m\), speed \(v_0\) and kinetic energy \(K=\tfrac12mv_0^2\) approaches a nucleus of charge \(Ze\) with impact parameter \(b\), the perpendicular distance from the nucleus to the line of the incident velocity, and leaves along a hyperbola with the same speed, deflected through the scattering angle \(\theta\) (Figure 1.2). Energy and the angular momentum \(L = mv_0b\) are conserved. It is convenient to define the head-on distance of closest approach \[d = \frac{1}{4\pi\varepsilon_0}\frac{zZe^2}{K} = 1.44\,\frac{zZ}{K\,[\text{MeV}]}\ \text{fm},\] using \(e^2/4\pi\varepsilon_0 = 1.44\) MeV fm.

(b) Impact parameter and cross-section

The momentum transfer has magnitude \(\lvert\Delta\mathbf p\rvert = 2mv_0\sin(\theta/2)\) and lies along the axis of symmetry of the orbit. Integrating the component of the force along that axis, \(\int F\cos\psi\,dt\), and changing the variable with \(mr^2\,d\psi/dt = mv_0b\) gives \(\lvert\Delta\mathbf p\rvert = (2zZe^2/4\pi\varepsilon_0v_0b)\cos(\theta/2)\). Equating the two expressions, \[\boxed{\,b = \frac{1}{4\pi\varepsilon_0}\frac{zZe^2}{2K}\cot\frac{\theta}{2} = \frac{d}{2}\cot\frac{\theta}{2}\,}\] so that distant collisions give small deflections and \(b\to0\) gives \(\theta\to180^\circ\).

Particles with impact parameters between \(b\) and \(b+db\) cross the ring of area \(d\sigma = 2\pi b\,\lvert db\rvert\) and emerge into the solid angle \(d\Omega = 2\pi\sin\theta\,d\theta\). With \(\lvert db/d\theta\rvert = d/[4\sin^2(\theta/2)]\), the differential cross-section \(d\sigma/d\Omega = (b/\sin\theta)\lvert db/d\theta\rvert\) becomes the Rutherford formula \[\boxed{\,\frac{d\sigma}{d\Omega} = \left(\frac{1}{4\pi\varepsilon_0}\frac{zZe^2}{4K}\right)^2\frac{1}{\sin^4(\theta/2)}\,}\] It predicts the \(\operatorname{cosec}^4(\theta/2)\) angular law and the dependences \(\propto Z^2\) and \(\propto1/K^2\propto1/v_0^4\), all confirmed by experiment; the divergence at \(\theta\to0\) reflects the infinite range of the Coulomb force and is removed by electron screening. The quantum-mechanical result for a Coulomb potential is identical. Note carefully that the number of particles per unit solid angle varies as \(\operatorname{cosec}^4(\theta/2)\), whereas the number per unit angle \(\theta\) contains the extra factor \(2\pi\sin\theta\): \[\boxed{\,\frac{dN}{d\theta}\propto\frac{\sin\theta}{\sin^4(\theta/2)}\propto\operatorname{cosec}^2\frac{\theta}{2}\cot\frac{\theta}{2}\,}\] Every particle with impact parameter smaller than \(b(\theta)\) is scattered beyond \(\theta\), so \(\sigma(>\theta) = \pi b^2 = (\pi d^2/4)\cot^2(\theta/2)\). For a foil of thickness \(t\) with \(n\) nuclei per unit volume, \(N_i\) incident particles give \[\boxed{\,dN = N_i\,n\,t\,\frac{d\sigma}{d\Omega}\,d\Omega,\qquad N(>\theta) = N_i\,n\,t\,\pi b^2\,}\] Hence for count ratios \(N\propto nt\,Z^2/[K^2\sin^4(\theta/2)]\).

(c) Distance of closest approach

In a head-on collision (\(b=0\), \(\theta = 180^\circ\)) the whole kinetic energy becomes Coulomb energy at the turning point, so \[\boxed{\,r_{\min}(b=0) = d = \frac{1}{4\pi\varepsilon_0}\frac{zZe^2}{K}\,}\] For \(b\neq0\) the velocity \(v_{\min}\) at closest approach is perpendicular to \(\mathbf r\), so angular momentum gives \(v_0b = v_{\min}r_{\min}\) and energy gives \(1 = b^2/r_{\min}^2 + d/r_{\min}\), a quadratic \(r_{\min}^2 - d\,r_{\min} - b^2 = 0\) with positive root \[\boxed{\,r_{\min} = \frac{d}{2} + \sqrt{\frac{d^2}{4} + b^2} = \frac{d}{2}\left(1 + \operatorname{cosec}\frac{\theta}{2}\right)\approx d + \frac{b^2}{d}\quad(b\ll d)\,}\] The Rutherford law holds only while \(r_{\min}\) exceeds the sum of the nuclear radii; at higher energies the nuclear force causes anomalous scattering, which gave the first estimates of nuclear size. Including target recoil replaces \(m\) by the reduced mass and \(K\), \(\theta\) by centre-of-mass values.

Example 1.1 A beam of 7.7 MeV \(\alpha\) particles falls on a gold foil (\(Z = 79\), \(n = 5.9\times10^{28}\) m\(^{-3}\)) of thickness 1.0 μm. Find (a) the impact parameter for \(\theta = 90^\circ\), (b) the fraction scattered beyond \(90^\circ\), (c) the count at \(60^\circ\) if a detector at