I performed the Rutherford experiment the other day, using Au thin foil of $2\,\mathrm{\mu{}m}$, a source/gun of α particles (241Am) and a detector/counter. α particles were shot from the source through a slit of $20\,\mathrm{mm} \times 1\,\mathrm{mm}$ (presumably attached to the foil, thus narrowing the effective area). In the first part of the experiment I was trying to measure the scattering angle without the foil for $0^\circ \ll |\theta| \ll 7^\circ$, where $\theta$ was the scattering angle and also the angle between the foil and the detector. For each angle I noted the number of counts and the time elapsed (in order to calculate the capacity=number of counts/time). The angles at which the capacity decreased by around 90% were $4^\circ$ and $-7^\circ$. Before I move on to describe the second part of the experiment using the foil, I'd like to pose a few questions as some things are not sufficiently clear to me.
Our booklet for this experiment instructs us to calculate $\mathrm{d}P_0(\theta)=\frac{n}{t}$ for each of the angles, then divide it for each angle by the angular width of the detector in order to determine $\frac{\mathrm{d}P_0}{\mathrm{d}\theta}$. If I understand the instructions correctly, $\frac{\mathrm{d}P_0}{\mathrm{d}\theta}$ is then apparently to be used to plot a Gaussian, which is to be integrated between $-\infty$ and $+\infty$ to yield the total $P_0$ (i.e. the "background" measurement, without the foil). Does that make sense? Normally it is the capacity itself which is plotted against the scattering angle to yield the Gaussian, and not the capacity divided by the angular width, isn't it? Furthermore, how exactly do I determine the angular width? The booklet indicates that in order to determine the angular width one must measure the width of the slit and the distance between the detector and the source/gun. However, I am not really sure I understand. I was under the impression that the angular width was simply given by $\Delta \theta=\Delta \Omega/(2\pi \sin\thetaθ)$, whereas $\Delta \Omega$ is the area of the slit used. Isn't it? I'd sincerely appreciate some feedback.