= Thin-foil scattering counts
{title2=$\mu=N_{\rm inc}\,n_s\,\eta\,\Delta\Omega\,\frac{d\sigma}{d\Omega}$}
Let $N_{\rm inc}$ be the number of projectiles actually incident on the foil during an exposure, $n_s$ its <foil areal number density>, and $\eta$ the probability of detecting an accepted particle. For a sufficiently small detector patch of <area> $A$ at distance $R$ from the scattering point, with its normal at <angle> $\psi$ to the line of sight,
$$
\Delta\Omega\simeq\frac{A\cos\psi}{R^2},\qquad
\mu(\theta)\simeq N_{\rm inc}\,n_s\,\eta\,\Delta\Omega\,
\frac{d\sigma}{d\Omega}(\theta).
$$
The product $n_s(d\sigma/d\Omega)\Delta\Omega$ is a dimensionless scattering probability in this approximation; $\mu$ is an <expected value> of counts, so it need not be an integer. The approximation requires a nearly constant <differential scattering cross-section> and efficiency across the patch and incident beam. For finite acceptance use $\mu=N_{\rm inc}n_s\int_{\rm detector}\eta(\Omega)(d\sigma/d\Omega)\,d\Omega$, including the incident beam's spread if appreciable. A screen facing the foil has $\psi=0$.
For an original numerical model, stipulate monoenergetic <alpha particles> with $z=2$, $E=5.00\,\mathrm{MeV}$, and a <gold> target with $Z=79$. Use the rounded model constant $e^2/(4\pi\varepsilon_0)=1.440\,\mathrm{MeV\,fm}$, where $1\,\mathrm{MeV}=10^6$ <electronvolts> and $1\,\mathrm{fm}=10^{-15}\,\mathrm m$. Thus $\kappa=227.52\,\mathrm{MeV\,fm}$ and $a=22.752\,\mathrm{fm}$. Choose $N_{\rm inc}=10^{11}$, $n_s=10^{21}\,\mathrm{m^{-2}}$, $\eta=1$, and $\Delta\Omega=10^{-3}\,\mathrm{sr}$. For example, $A=1\,\mathrm{mm^2}$ facing the foil at $R\simeq31.6\,\mathrm{mm}$ gives this small-patch <solid angle>.
At $\theta=90^\circ$, $b=a$ and $d\sigma/d\Omega=a^2=517.654\,\mathrm{fm^2\,sr^{-1}}=5.17654\,\mathrm{b\,sr^{-1}}$, using $1$ <barn> $=100\,\mathrm{fm^2}=10^{-28}\,\mathrm{m^2}$. The expected signal is $\mu\simeq51.8$ counts. At $60^\circ$ it is four times larger, about $207$ counts. Doubling <foil areal number density> or detector <solid angle> doubles these expected signals as long as the approximations continue to hold.
\Image[/rutherford-angular-cross-section-and-counts.png]
{title=Calculated Rutherford angular cross-section and finite-detector counts}
{description=Original calculation for the stated ideal monoenergetic alpha beam and thin gold foil. The curve is a differential cross-section; the bars are expected counts for the specified small detector solid angle at separate settings. They are not measured 1909 or 1913 data, and do not sum to a total scattering probability. The curve is shown only from 20 to 160 degrees.}
An observation also contains backgrounds. For a stable source and matched geometry, an estimator is $C_{\rm net}=C_{\rm foil}-(T_{\rm foil}/T_{\rm blank})C_{\rm blank}$, where $T$ denotes exposure duration. If source activity changes, normalize by incident exposure as well. <Background subtraction> removes an estimated background; it does not make finite counting fluctuations disappear.
Back to article page
