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Rutherford Scattering

  1. Dec 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculate the cross section for the scattering of a 10 MeV alpha particle by a gold nucleus (Z=79, A=197) through an angle greater than a) 10 degrees b) 20 degrees c) 30 degrees. Neglect Nuclear recoil.

    2. Relevant equations
    http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/crosec.html

    3. The attempt at a solution
    I've seen a number of equations that relate to rutherford scattering. The one on hyperphysics looks neat, but it doesn't say what the variables in the equation are. I don't know what the small 'ke' means and what the two angles are supposed to be in the equation either. I'm assuming they are both the scattering angle.
     
  2. jcsd
  3. Dec 29, 2015 #2
    You can find that by dimension check.
     
  4. Dec 29, 2015 #3
    small ke is Newton / metre? I assume it has something to do with the Coulomb Force / Impact Parameter.
     
  5. Dec 29, 2015 #4
    $$\frac{k\,e^2}{KE}$$
    must be on meters.
     
  6. Dec 29, 2015 #5
    kg/s2 on the numerator would give units of barns
     
  7. Dec 29, 2015 #6
    ahhhh! I was thinking in the wrong units. So I found that ke2 = 1.44 MeV*fm

    If I plug that into the cross section equation given by hyperphysics I get an answer of 0.022813 fm2

    But my textbook says the answer is 5.3*104 fm2
     
  8. Dec 29, 2015 #7

    jtbell

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    Staff: Mentor

    Try converting the given kinetic energy (KE) into joules. Then you can use standard SI (MKS) units for the rest. k is the Coulomb constant from Coulomb's Law: k = 1/(4πε0). e is the elementary charge (magnitude of charge on electron or proton).
     
    Last edited: Dec 29, 2015
  9. Dec 29, 2015 #8
    I still get the incorrect answer.

    k= 8.98×109
    e = 1.60x10-19

    I use those values and plug it in to the equation and I don't get 5.3*104 fm2
     
  10. Dec 29, 2015 #9

    jtbell

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    Staff: Mentor

    What did you use for the KE? (in J)
     
  11. Dec 29, 2015 #10
    1.60218*10-12
     
  12. Dec 29, 2015 #11

    jtbell

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    Staff: Mentor

    Your input numbers agree with mine, but I do get the textbook's answer (for 10°). Check your arithmetic again, and if you still can't find the error, show exactly how you set it up, with the numbers substituted into the equation.

    Make sure your calculator is set to use degrees instead of radians for trig functions.
     
  13. Dec 29, 2015 #12
    π * 22 * ( 8.9*109 * ((1.6*10-19)2) / 1.6*10-12 ) (1+cos(10) / 1-cos(10) )

    = 3.3*10-29

    Sorry about the formatting, but I've got it in degree mode and I've checked all the brackets etc.
     
  14. Dec 29, 2015 #13
    And my units seem right with dim. analysis

    N*m2/C2 * C2 / kg * m2*s-2
     
  15. Dec 29, 2015 #14

    jtbell

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    Staff: Mentor

    Z is for gold (79), not the alpha particle (2).

    Also, you omitted one power of 2. Look at the original equation carefully.

    Finally, you may have some discrepancy from roundoff error because you used only two significant figures.
     
  16. Dec 29, 2015 #15
    oh, damn! I should've seen that. I tend to look at equations and hope that someone has wrote, where Z= ...., k=...., because sometimes I either forget things, i.e. like coulomb constant and elementary charge, and I find it easier to understand relationships with equations instead of a slab of text. Thanks for your help! :)
     
  17. Dec 29, 2015 #16
    adding the power of 2 and changing the Z to 79, i get 5.19e-26
     
  18. Dec 29, 2015 #17
    I'm still out by some order of magnitude, but I'm not sure why.
     
  19. Dec 29, 2015 #18
    my answer is in m, and not m2. If I square my answer though it's out by an even greater magnitude...

    textbook = 5.3*104 fm2

    me = 5.19 *10-26 m = 5.19*10-11 fm
     
  20. Dec 29, 2015 #19
    Figured it out. I replaced the units with MeV and got the correct answer.
     
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