Rutherford Experiment - practical and theoretical questions

Question

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.

Answer 1

I'm going to answer your question back-to-front:

I was under the impression that the angular width was simply given by Δθ=ΔΩ/(2πsinθ), whereas ΔΩ is the area of the slit used. Isn't it?

Not quite, $\Delta\Omega$ represents the solid angle subtended by the scattered particles, but the relation is correct. However, I think that for your experiment you don't need this relation right now. What your booklet asks you to measure is $\frac{dP_0}{d\theta}$ so let's stick with that.

First things first, make measurements $P_0$ as a function of the angle $\theta$. Plot your data with $P_0$ on the vertical axis vs $\theta$ on the horizontal axis. Next, connect your data points with straight lines (which all have different gradients). Next, measure the gradient of each line i.e measure $\frac{\Delta P_0}{\Delta\theta}$ between each point, noting the range $\theta_{min}$ and $\theta_{max}$ for each value of $\frac{\Delta P_0}{\Delta\theta}$. ($\frac{\Delta P_0}{\Delta\theta}$ is our approximation of $\frac{dP_0}{d\theta}$).

Now plot a new graph as follows. Plot your measured values of $\frac{\Delta P_0}{\Delta\theta}$ as being constant in the range $\theta_{min}$ to $\theta_{max}$.

If I understand the instructions correctly, dP$_0$/dθ is then apparently to be used to plot a Gaussian, which is to be integrated between -∞ and +∞ to yield the total P$_0$ (i.e. the "background" measurement, without the foil). Does that make sense?

I don't think this is correct. $\frac{dP_0}{d\Omega}$ varies as $\frac{1}{sin^4\theta}$ (see this: http://en.wikipedia.org/wiki/Rutherford_scattering). From this its not so obvious that $\frac{dP_0}{d\theta}$ would be Gaussian-shaped. As for the total $P_0$ (in some range of $\theta$, this is simply the integral of your "$P_0$ as a function of the angle $\theta$" distribution that you have measured.