= Solution
For elastic scattering by the <Yukawa potential>, take $\kappa>0$ and use the <first Born approximation> in the plane-wave normalization
$$
f_B(\mathbf q)=-\frac{m}{2\pi\hbar^2}\int e^{-i\mathbf q\cdot\mathbf r}V(r)\,d^3r.
$$
Choose the polar axis along the <momentum transfer> $\hbar\mathbf q$. The angular integral is
$$
\int e^{-iqr\cos\vartheta}\,d\Omega
=2\pi\int_{-1}^{1}e^{-iqru}\,du
=4\pi\frac{\sin(qr)}{qr}.
$$
The radial volume element cancels the remaining powers of $r$, leaving
$$
f_B(q)=-\frac{2mA}{\hbar^2q}\int_0^\infty e^{-\kappa r}\sin(qr)\,dr
=-\frac{2mA}{\hbar^2(q^2+\kappa^2)}.
$$
For real $A$, the <differential scattering cross-section> is therefore
$$
\boxed{\frac{d\sigma}{d\Omega}=|f_B(q)|^2
=\left[\frac{2mA}{\hbar^2(q^2+\kappa^2)}\right]^2}.
$$
This also has a finite forward limit when $\kappa>0$. The <Born amplitude for a Yukawa potential> provides a screened route to the <Coulomb potential> by taking $\kappa\to0^+$ at fixed $q>0$.
For an <alpha particle> of charge $2e$ scattering from a heavy nucleus of charge $Ze$, set $A=2Ze^2/(4\pi\epsilon_0)$. Neglect recoil, so that the reduced mass is approximately the projectile mass, $E=\hbar^2k^2/(2m)$, and elastic geometry gives
$$
q^2=|\mathbf k_{\rm out}-\mathbf k_{\rm in}|^2
=4k^2\sin^2\frac\theta2.
$$
The Coulomb-limit <scattering amplitude> becomes
$$
f_B(\theta)=-\frac{A}{4E\sin^2(\theta/2)}
=-\frac{Ze^2}{8\pi\epsilon_0E\sin^2(\theta/2)},
$$
which gives the <Rutherford differential cross-section>
$$
\boxed{\frac{d\sigma}{d\Omega}
=\left[\frac{Ze^2}{8\pi\epsilon_0E\sin^2(\theta/2)}\right]^2}.
$$
The unscreened <Coulomb potential> has long range, so taking this limit does not justify the <Born approximation> for its full scattering state, particularly for a strongly charged target. The exact <Rutherford scattering> amplitude differs by a phase and has the same modulus. Thus the resulting <differential scattering cross-section> remains applicable to the ideal nonrelativistic Coulomb problem beyond the weak-potential regime.
To relate the detector counts to the <atomic number>, first compute the number of target atoms per illuminated area. With molar mass $M=0.197\ \mathrm{kg\,mol^{-1}}$, the <number density> integrated through the foil is
$$
\mathcal N=\frac{\rho tN_A}{M}
=\frac{(19.3\times10^3)(2.1\times10^{-7})(6.02214\times10^{23})}{0.197}
=1.239\times10^{22}\ \mathrm{m^{-2}}.
$$
The detector's transverse area is already its area normal to the scattered ray. Its small <solid angle> is
$$
\Delta\Omega\simeq\frac{10^{-6}}{(10^{-2})^2}=10^{-2}\ \mathrm{sr}.
$$
There is no extra $\cos45^\circ$ factor: the foil is traversed normally and the detector area is transverse to its line of sight.
For a thin foil with negligible <multiple scattering> and <energy> loss, the detected fraction is $f=\mathcal N(d\sigma/d\Omega)\Delta\Omega$. Thus the measurements imply
$$
\frac{d\sigma}{d\Omega}
=\frac{3.7\times10^{-7}}{(1.239\times10^{22})(10^{-2})}
=2.986\times10^{-27}\ \mathrm{m^2\,sr^{-1}}.
$$
Using $E=7.68\times10^6e\ \mathrm J=1.23047\times10^{-12}\ \mathrm J$ in the <Rutherford differential cross-section> gives
$$
Z=\frac{8\pi\epsilon_0 E\sin^2(22.5^\circ)}{e^2}
\sqrt{\frac{f}{\mathcal N\Delta\Omega}},\qquad
\boxed{Z\simeq85.4\approx85}.
$$
\b[The stated measurements do not reproduce gold's <atomic number> 79 in this model.] The values and units above agree with the original PDF. Substituting $Z=79$ instead would predict $f\simeq3.17\times10^{-7}$ with the other inputs unchanged, rather than $3.7\times10^{-7}$. No stated correction accounts for that discrepancy; forcing the answer to 79 would require changing the data or adding an experimental correction not specified in the problem.
Back to article page
