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Rutherford experiment

  1. Feb 14, 2014 #1
    Hi,

    My question is about the famous paper by E. Rutherford on the scattering of alpha particles (1911). The paper is easy to find on internet. Rutherford gives the formula for the electric field inside an atom, at a distance r from the nucleus (here reduced to a point):

    X = N e (1/r^2 - r/R^3)

    Here N e in the electric charge in the nucleus and R is the radius of the atom. After he says

    It is not difficult to show that the deflection (supposed small) of an electrified particle due to this field is given by

    theta = b/p (1 - p^2/R^2)^(3/2)

    where p is the perpendicular from the center on the path.

    I can obtain X but I don't see where the formula for theta comes from.

    Any suggestion?

    Thanks DH
     
  2. jcsd
  3. Feb 14, 2014 #2

    BvU

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    If I were old Ernest, I would integrate the vertical component of the force over the section of the trajectory through the nucleus. If theta is small enough, pretending it is straight may be OK, and it's easier.
     
  4. Feb 15, 2014 #3
    Hi,
    Thanks for the answer.
    Let the trajectory of the alpha particle be on the x-axis.

    Are you suggesting to calculate

    \int_{t1}^{t2} Fy dt

    where Fy is the y-coordinate of the force?

    DH
     
  5. Feb 15, 2014 #4

    BvU

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    I found Rutherford-1911 where your formulas feature. There also is an explanation of what b stands for, which you would have included in your problem statement if you would have used the template. Please use it from now on. PF has a simple rule: no template, no assistance. It would have saved me some time that I could have used for others. Now I have to do some errands, so I am short on time.

    But yes, (read: atom in my post, the nucleus is considered pointlike, somewhat incongruent in this context: it still had to be discovered. Better to speak of the center of the atom, but never mind). t1 and t2 can be related to R and p and the speed of the ##\alpha##. So you change from dt to dx. Some ##\beta## comes in with ## \sin\beta = {x\over R}##; perhaps you go from dx to d##\beta##. All constants go into b and there you are!
     
    Last edited: Feb 15, 2014
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