07 Jul 2026

Alpha-Particle and Rutherford Scattering

Coulomb orbit of an alpha particle, impact parameter, closest approach, Rutherford cross section, and laboratory-frame conversion.

msc semester-ii quantum-mechanics scattering alpha-scattering rutherford-scattering coulomb-potential

An alpha particle has charge $Z_1e=2e$. A nucleus of charge $Z_2e=Ze$ repels it through

\[V(r)=\frac{\kappa}{r}, \qquad \kappa=\frac{Z_1Z_2e^2}{4\pi\varepsilon_0}>0.\]

In the centre-of-mass frame, use the reduced mass $\mu$, asymptotic speed $v$, energy $E=\tfrac12\mu v^2$, and angular momentum $L=\mu vb$.

Repulsive Coulomb orbit

The radial equation of motion and angular-momentum conservation are

\[\mu(\ddot r-r\dot\phi^2)=\frac{\kappa}{r^2}, \qquad L=\mu r^2\dot\phi.\]

Set $u=1/r$ and let a prime denote $d/d\phi$. Since

\[\dot\phi=\frac{L}{\mu}u^2,\]

the chain rule gives

\[\dot r =\frac{d(1/u)}{d\phi}\dot\phi =-\frac{u'}{u^2}\frac{L}{\mu}u^2 =-\frac{L}{\mu}u',\]

and

\[\ddot r =-\frac{L}{\mu}u''\dot\phi =-\frac{L^2}{\mu^2}u^2u''.\]

The angular term is

\[r\dot\phi^2 =\frac1u\left(\frac{L}{\mu}u^2\right)^2 =\frac{L^2}{\mu^2}u^3.\]

Substitution in the radial equation, followed by multiplication by $\mu/(L^2u^2)$, produces

\[\frac{d^2u}{d\phi^2}+u=-\frac{\mu\kappa}{L^2} =-\frac1p, \qquad p=\frac{L^2}{\mu\kappa}.\]

The homogeneous solutions are $\cos\phi$ and $\sin\phi$, while $-1/p$ is a particular solution. Combining the sine and cosine into one shifted cosine,

\[u=\frac1p\left[e\cos(\phi-\phi_0)-1\right],\]

and therefore

\[\boxed{ r(\phi)=\frac{p}{e\cos(\phi-\phi_0)-1}. }\]

The energy relation follows directly from

\[\dot r=-\frac{L}{\mu}\frac{du}{d\phi}, \qquad E=\frac{L^2}{2\mu} \left[ \left(\frac{du}{d\phi}\right)^2+u^2 \right]+\kappa u.\]

Write $\chi=\phi-\phi_0$. Since

\[u=\frac{e\cos\chi-1}{p}, \qquad u'=-\frac{e\sin\chi}{p},\]

and $1/p=\mu\kappa/L^2$, the dimensionless numerator inside the energy is

\[\begin{aligned} &e^2\sin^2\chi+(e\cos\chi-1)^2 +2(e\cos\chi-1)\\ &\hspace{3em}=e^2-1. \end{aligned}\]

Therefore

\[E=\frac{\mu\kappa^2}{2L^2}(e^2-1).\]

Thus

\[e^2=1+\frac{2EL^2}{\mu\kappa^2}.\]

With $L=\mu vb$, $2E=\mu v^2$, and

\[a=\frac{\kappa}{2E},\]

this becomes

\[\boxed{ e=\sqrt{1+\frac{b^2}{a^2}}, \qquad p=\frac{b^2}{a}. }\]

The curve is the exact hyperbolic solution of the repulsive Coulomb orbit. Its asymptotes, impact parameter, deflection angle, and perpendicular periapsis tangent use the same \(a\) and \(b\).

The denominator vanishes on the two asymptotes. Choose the incoming asymptote at $\phi=\pi$ and the outgoing one at $\phi=\theta$. Symmetry about the periapsis fixes

\[\phi_0=\frac{\pi+\theta}{2}.\]

At either asymptote,

\[\frac1e =\cos\frac{\pi-\theta}{2} =\sin\frac{\theta}{2}.\]

Combining $e^2=\csc^2(\theta/2)$ with $e^2=1+b^2/a^2$,

\[\cot^2\frac{\theta}{2}=\frac{b^2}{a^2},\]

so the positive impact parameter is

\[\boxed{ b=a\cot\frac{\theta}{2} =\frac{\kappa}{2E}\cot\frac{\theta}{2}. }\]

At the periapsis $\phi=\phi_0$, $dr/d\phi=0$. Its distance from the nucleus is

\[r_{\min}=\frac{p}{e-1} =\frac{a(e^2-1)}{e-1} =a(e+1),\]

or

\[\boxed{ r_{\min}=a+\sqrt{a^2+b^2}. }\]

For a head-on collision, $b=0$ and $r_{\min}=\kappa/E$.

Rutherford differential cross section

Differentiate the deflection law:

\[\frac{db}{d\theta} =-\frac{a}{2}\csc^2\frac{\theta}{2}.\]

Using $\sin\theta=2\sin(\theta/2)\cos(\theta/2)$ in $d\sigma/d\Omega=(b/\sin\theta)|db/d\theta|$,

\[\begin{aligned} \frac{d\sigma}{d\Omega} &=\frac{ a\cot(\theta/2) }{ 2\sin(\theta/2)\cos(\theta/2) } \frac{a}{2}\csc^2\frac{\theta}{2}\\ &=\frac{a^2}{4}\csc^4\frac{\theta}{2}. \end{aligned}\]

Therefore

\[\boxed{ \frac{d\sigma}{d\Omega} =\left(\frac{\kappa}{4E}\right)^2 \csc^4\frac{\theta}{2}. }\]

The forward divergence makes the ideal Coulomb total cross section infinite. With a lower acceptance angle $\theta_{\min}$,

\[\begin{aligned} \sigma(\theta\geq\theta_{\min}) &=2\pi\int_{\theta_{\min}}^\pi \frac{a^2}{4}\csc^4\frac{\theta}{2} \sin\theta\,d\theta\\ &=2\pi a^2 \int_{\theta_{\min}/2}^{\pi/2} \frac{\cos x}{\sin^3x}\,dx. \end{aligned}\]

Here $x=\theta/2$, and

\[\int\frac{\cos x}{\sin^3x}\,dx =-\frac{1}{2\sin^2x}.\]

Evaluation at the limits produces

\[\boxed{ \sigma(\theta\geq\theta_{\min}) =\pi a^2\cot^2\frac{\theta_{\min}}{2}. }\]

These formulas use $E=E_{\mathrm{cm}}$. For a target initially at rest,

\[E_{\mathrm{cm}} =\frac{m_2}{m_1+m_2}E_{\mathrm{lab}}.\]

For a non-heavy target, the measured angle also requires

\[\tan\theta_{\mathrm{lab}} =\frac{\sin\theta_{\mathrm{cm}}} {\cos\theta_{\mathrm{cm}}+m_1/m_2},\]

and the laboratory cross section is obtained with the corresponding solid-angle Jacobian. The common heavy-nucleus approximation $m_2\gg m_1$ makes both corrections small.

The orbit equation, asymptotes, periapsis, deflection law, and cross sections are verified in the Maxima worksheet.