Rutherford experiment practical questions.

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Hi,

Homework Statement


I performed the Rutherford experiment the other day, using Au thin foil of 2μm, a source/gun of α particles (241Am) and a detector/counter. α particles were shot from the source through a slit of 20mmX1mm (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°≤|θ|≤7°, where θ 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° and -7°.
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.


Homework Equations





The Attempt at a Solution


Our booklet for this experiment instructs us to calculate dP0(θ)=n/t for each of the angles, then divide it for each angle by the angular width of the detector in order to determine dP0/dθ. If I understand the instructions correctly, dP0/dθ is then apparently to be used to plot a Gaussian, which is to be integrated between -∞ and +∞ to yield the total P0 (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 Δθ=ΔΩ/(2πsinθ), whereas ΔΩ is the area of the slit used. Isn't it?
I'd sincerely appreciate some feedback.
 

Answers and Replies

  • #2
Simon Bridge
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Hi,

Homework Statement


I performed the Rutherford experiment the other day, using Au thin foil of 2μm, a source/gun of α particles (241Am) and a detector/counter. α particles were shot from the source through a slit of 20mmX1mm (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°≤|θ|≤7°, where θ was the scattering angle and also the angle between the foil and the detector.
Usually one would place the detector at a specific angle and then count the number of alpha particles detected in a set period of time.

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° and -7°.
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 dP0(θ)=n/t for each of the angles, then divide it for each angle by the angular width of the detector in order to determine dP0/dθ. If I understand the instructions correctly, dP0/dθ is then apparently to be used to plot a Gaussian, which is to be integrated between -∞ and +∞ to yield the total P0 (i.e. the "background" measurement, without the foil).

Lets say the detector has a circular hole with a hole area A.
Alpha particles are scattered to all angles - what the detector does is look only at the ones that are scattered through the hole, and you count them.

The detector picks up all the alphas that are scattered into the range of angles that would hit the hole in the detector.

If you did N/AT, then that would be the area-density of the particle-rate ... which would be the particle flux density.

What you want is the angle density.

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?
No. You can do it that way, but it is harder to compare to the theory.
BTW: the distribution need not be a Gaussian ... all you are doing is plotting a graph.
Decide what shape the graph is after it is plotted.

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 Δθ=ΔΩ/(2πsinθ), whereas ΔΩ is the area of the slit used. Isn't it?
You can see what they mean by sketching the setup ... you treat the slit as the apex of a triangle and the diameter of the hole as the base.
The angular width is the apex angle ... this gets smaller the further away the hold is from the slit.
Draw it and see.

The formula you gave is for a different situation.
 
  • #3
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Okay, my problem is currently slightly different. I have obtained a seemingly exorbitant number of counts per second for the total P0. I have created a Gaussian of dP0/Δθ(θ), where P0 is the number of counts of alpha particles through a 1.05X19.86 mm slit divided by the time elapsed and Δθ(θ) is the angular width of the detector. At this stage the Au foil was not used, as this was merely the "background" measurement. I have integrated that Gaussian wrt θ between −∞ and +∞ to obtain the total P0. I have done all that following my instructor's advice! Supposedly he knew what he was talking about. However, this yielded a total P0 equal to 24,000 counts/sec, which seems exorbitant and wrong. I'd be indebted for any help trying to figure out where the mistake is!
When used further in the experiment that P0 seems to yield a wrong value for the expected dσ/dΩ. Moreover, it is unlikely that without the Au leaf and at very small angles the number of counts/second would be that high! Let's examine for instance the readings I obtained for the angle θ=0: number of counts - 1072; time elapsed - 14.1 sec; P0=1072/14.1=76.028 [1/sec]; Δθ = width of slit/distance between source and detector = 1.05mm/44mm = 0.02386; P0/Δθ = 3186.436 [1/sec]. There seems to be something wrong here. Possibly with the evaluation of Δθ.
As mentioned, I'd sincerely appreciate some insight on this.
 
Last edited:
  • #4
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As an angle, θ is limited, but given the small beam spot this should not matter.

However, this yielded a total P0 equal to 24,000 counts/sec, which seems exorbitant and wrong.
That looks too high, given the values you show afterwards.

Moreover, it is unlikely that without the Au leaf and at very small angles the number of counts/second would be that high!
Higher than with Au, as there is no Au scattering atoms out of their path.

P0/Δθ = 3186.436 [1/sec]
That could be true, and if you integrate your function correctly it will give a total rate significantly below 3186/sec.
 
  • #5
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The integration as far as I am concerned was correct. It was carried down by Wolfram's engine and the Gaussian's mu and sigma were provided by MATLAB. Where's my error then?
 
  • #7
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I did use degrees, but made sure I was consistent all along in my conversions and calculations etc. I mean, whenever necessary I took meticulous care to multiply the values by pi and divide by 180 or vice versa. Any other suggestions? I'd also like to post a few of my graphs, which might provide further insight. Would you agree to have a look?
 
  • #8
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I did use degrees, but made sure I was consistent all along in my conversions and calculations etc. I mean, whenever necessary I took meticulous care to multiply the values by pi and divide by 180 or vice versa. Any other suggestions? I'd also like to post a few of my graphs, which might provide further insight. Would you agree to have a look?
This is in radians:
Δθ = width of slit/distance between source and detector = 1.05mm/44mm = 0.02386
In degrees, it is larger by a factor of 180/pi.
 
  • #9
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You were absolutely right. I have now corrected that and obtained 413 counts/second! However, when I am trying to fit my data to the theoretical formula, I obtain the plot in the attachment. Which raises two questions:
(1) I am not supposed to take the absolute value of all the angles, right?
(2) I cannot figure out why the plot won't reach the top most datum. I am limiting the range to -0.7 to -0.174, but that should still suffice. The model I am using for plotting the graph is of course the theoretical formula, namely:
[(zαzAue2)/(4Eeff)]2*(sin-4(θ/2)) = 1.6763*10-22*(sin-4(θ/2)) mm2, for zAu=79, zα=2, Eeff=4.4Mev
I am trying to compare between the two formulae and obtain a minimum chi squared to indicate the degree of congruity. Is it okay that the top datum is not intersected, or is there a better way to approach this?
 

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  • #10
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NB The graph represents the readings registered using the Au leaf (i.e. the second part of the experiment)
 
  • #11
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You were absolutely right. I have now corrected that and obtained 413 counts/second! However, when I am trying to fit my data to the theoretical formula, I obtain the plot in the attachment. Which raises two questions:
(1) I am not supposed to take the absolute value of all the angles, right?
Probably not.

(2) I cannot figure out why the plot won't reach the top most datum.
The model is not perfect, I think. Extend the plotting range a bit and it could look better.

I am trying to compare between the two formulae and obtain a minimum chi squared to indicate the degree of congruity. Is it okay that the top datum is not intersected, or is there a better way to approach this?
Could work, but does not have to.
 
  • #12
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Even when I extend the range it still doesn't intersect that top most value. Any idea why?
 
  • #13
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Unless I render it continuous, which is not what's desired.
 
  • #14
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Even when I extend the range it still doesn't intersect that top most value. Any idea why?
A fit function does not have to intersect everything, if it is close to that value it is probably fine. Is the angular difference between data and fit function within the uncertainty (or 2-3 times this value) for your angle?
 
  • #15
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I am not sure I understand what you mean. Would you please try to clarify?
 
  • #16
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An example "fit" (actually made with paint).
The black crosses are data points with uncertainties, the red line is the fit, the blue points mark the "actual" values corresponding to the rightmost two data points. They are close to the measured points, even if they are far above the fit function.

attachment.php?attachmentid=65460&stc=1&d=1389311728.jpg



What is your precision for those angles in your data points?
 

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  • #17
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Alright, I have managed with this one. By the way, why is such a poor chi square reduced obtained for the Gaussian fit? In other words, for what reasons does dP0/Δθ(θ) poorly fit the Gaussian model? Is it because the number of readings is not sufficient for the CLT to hold?
 
  • #18
Simon Bridge
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Either that or it is just not a great fit to a gaussian.
Remember why you are doing the test in the first place?
 
  • #19
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Foremost, and perhaps merely, in order to determine the value of the "background" scattering, the total P0. But why would that necessarily require a Gaussian and not a different model? Is it because a Gaussian is easily integrated (withal converges) between infinite boundaries?
 
  • #20
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But why would that necessarily require a Gaussian and not a different model?
It does not require a Gaussian, and it is not a Gaussian - but this is a good approximation, and a Gaussian is a relatively easy function.
 
  • #21
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Would you say that perhaps, considering the low number of measurements, a Poisson distribution used as a model to calculate the total P0 might have yielded better/more accurate results?
 
  • #22
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I don't see how you plan to use a Poisson distribution here.
 
  • #23
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I don't plan on using anything else than that Gaussian. I am simply wondering what might have improved the approximation of the total capacity. The Poisson distribution provides a robust model for a relatively low number of counts per time elapsed; I figured it might do a better job than the Gaussian here. Any other suggestion then?
 
  • #24
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The Poisson distribution provides a robust model for a relatively low number of counts per time elapsed
The Poisson distribution of what?
Sure each datapoint follows a Poisson distribution, but the angular distribution does not, and you just have a single value per angle - nothing to base a statistical analysis on.
 
  • #25
Simon Bridge
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I am simply wondering what might have improved the approximation of the total capacity. ... Any other suggestion then?
I wouldn't bother in your position - yes you could come up with a different model that may help get a closer approximation but how would you tell? You are only doing the one experiment.

So far you have a distribution that is sort-a bell shaped and you've found, at least, that it is not as well approximated by a gaussian as you'd have hoped for. Nothing wrong with that - if you look into it more deeply you may be able to figure out how this impacts the background estimate - and how much that affects the end result and the aim.

Whatever - soldier on.
 

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