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.

  • 1
    $\begingroup$ This would be a lot easier to pick apart with a good diagram of your setup. You are referring to a lot of different length and widths and other measurements, and we have to be very sure we know what each one is before we can answer or certain. $\endgroup$ Jan 4, 2014 at 17:43
  • $\begingroup$ Additionally you need to be very sure you understand what quantity(ies) you are trying to measure and what the represent physically. If you are trying to get the differential cross-section you need to be absolutely clear on what one of those is. $\endgroup$ Jan 4, 2014 at 17:46

1 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.

  • $\begingroup$ Second, let's start from the beginning of your answer, and also the beginning of the data analysis. The booklet explicitly states that dP is to be divided by d$\theta$ for each angle and that d$\theta$ is to be calculated via measuring the distance between the source/gun and the detector as well as the width of the slit. Why? Could you make sense of all of this, please? $\endgroup$
    – peripatein
    Jan 4, 2014 at 17:32
  • $\begingroup$ Ah yes I see. In order to measure the outgoing angle $\theta$ one assumes that the ingoing $\alpha$s are incident to the foil at $0\degrees$, which is why there is a slit in front of your foil. However as the slit has some width, this means that the incoming $\alpha$s have some initial angular distribution (assumed to be uniform) and simply described by the geometry of the system (ie the width of the slit and the distance to the source). You can use this incoming angular spread to estimate the corresponding uncertainty in the outgoing $\theta$ distribution. (going to be continued) $\endgroup$
    – kd88
    Jan 4, 2014 at 18:14
  • $\begingroup$ This makes a lot of sense. Alright, so by what value should I divide my n/t each time? And why couldn't that value be determined simply by Δθ=ΔΩ/(2πsinθ)? Where ΔΩ is the area of the slit. $\endgroup$
    – peripatein
    Jan 4, 2014 at 18:17
  • 1
    $\begingroup$ but ΔΩ isn't the area of the slit, it is the spread of the solid angle of the scattered $\alpha$ particles. This is different from the incoming spread of the solid angle. I will edit my answer shortly now I understand your question, but I would ignore my original answer for the moment and use the comments that I have provided subsequently. $\endgroup$
    – kd88
    Jan 4, 2014 at 18:35
  • 1
    $\begingroup$ Note that the area of the slit divided by $4\pi r^2$ for $r$ the distance from slit to foil is roughly the proper $\Delta \Omega$. Not precisely, unless both the slit and the foil are small, but pretty close. $\endgroup$ Jan 4, 2014 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.