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Rutherford Data Analysis

  1. Dec 17, 2007 #1

    Pythagorean

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    1. The problem statement, all variables and given/known data

    This is for a lab class, I'm writing a report and giving a presentation. Tomorrow is the day and I've just received the final remarks on my lab writeup, most of which are simple and obvious enough, but this one really bugs me: I think my teacher is wrong.

    Rutherford's Equation for scattering alpha particles on the nucleus is:

    [tex]N(\theta)=\frac {K} {sin^4 (\frac {\theta} {2}})[/tex]

    we can't compare this function directly to our data since it goes about the 0-angle, which leads to infinite, so we log transform it (here, I've assumed natural logarithm, but have since tried log base 10 just to be sure)

    [tex]\log N = log K - 4log(sin(\frac{\theta} {2}) )[/tex]

    So I plot logN vs log(sin(theta/2)) to verify a slope of -4. I don't quite get 4 (I get -3.5 for gold and -4.4 for silver), and I don't get a straight line, and in the red ink responses, there's a lot of "no! not in radians! do it in degrees! so that you won't get the bending at the edges!".

    3. The attempt at a solution

    I don't agree with my professor. I think the sin of theta in degrees or radians should be the same. Perhaps I am misunderstanding him. I made the mistake of putting units of 'radians' on my plot of the logarithmic transform of Rutherford's equation. I realize now the plot should be unitless. I'm wondering if I confused him with this, or whether he's making a point I'm missing.

    Thank you for your time.
     
    Last edited: Dec 17, 2007
  2. jcsd
  3. Dec 17, 2007 #2

    Pythagorean

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    the plot for gold if you're interested:

    [​IMG]
     
  4. Dec 18, 2007 #3

    malawi_glenn

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    I dont think he is correct, I agree with you. Sin(180) = Sin(pi) etc, it doesnt matter. If he think it matters, he must demonstrate this mathematically. But it is customary to have the scattering angle in degrees, in the plot (axis-labels).

    The things is that you should not get a totaly straight line, due to the fact that a nucleus has shape, Rutherford is just for point particle, you must include a formfactor to get better (but not optimal) fit to data.

    Anyway, go to him and make him demonstrate that to you, that Sin should be in degrees.
     
  5. Dec 18, 2007 #4

    Pythagorean

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    Yeah, I straightened him out :P For about an hour there, I thought I that maybe I had missed something all these years playing with trig functions. It's funny how much faith we put into professors, sometimes.

    Anyway, thanks for pointing out the scattering equation is based on a point particle. That might come in handy when I present this tomorrow.
     
  6. Dec 18, 2007 #5

    malawi_glenn

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    Yes, I often trust my professors much. But in the end, I must go and say that they had wrong in their lecture notes etc, and they are just happy when someone points out their misstakes:) Thats also what differs from a good and less good student. A good student trust himselft to a higher degree and see the misstakes the teacher does. I had a professor who get a bit "angry" when nobody in the class room points out his misstakes and asks questions. Thats a way to see that the students are following :)

    Yeah, you can find the concept of formfactor in almost any nuclear physic book. It is just the fourier transform of the charge distribution function. Then the total differential cross section becomes:

    [tex] N(\theta ) = N(\theta )_{Ruth} \cdot \vert F(\theta ) \vert ^2 [/tex]

    where [itex] F(\theta ) [/itex] is the form factor.

    When using form factor, one can insert different models of the charge distribution (which is very similar to the nucleon distribution, protons and neutrons are belived to be very homogenously distributed in the nucleus) and playing with their parameters. For example one can model the nucleus as a square well, gaussian, fermi distribution etc. =)

    Good luck!
     
    Last edited: Dec 18, 2007
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