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Rutherford scattering

  1. Oct 7, 2007 #1

    uae

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    hi,

    i am a new student here in this forums from uae and i am studing in uaeu my majore is general physics >>>

    in this corse i have problem in nuclier physics in rutherford scattering >>>
    can you help ma pleaseeeee the problem are :

    firest one is :



    Show that the distance of closest approach d, in Rutherford scattering leading to an angle of deflection Θ ,is given by :




    [​IMG]


    d=p\2(1+cosecΘ\2)
    Where p is defined in fig .
    (use the conservation of energy and angular momentum)


    secound one is :



    Show that the Rutherford differential scattering cross-section formula can be written in terms of the squared momentum transfer q^2 as
    Dσ\dq^2=4πZzα^2(ЋC)^2\q^4ν^2
    Where αis the fine – structure constant and v is velocity of the deflected particle



    thank you for all in this forums
     
  2. jcsd
  3. Oct 7, 2007 #2

    malawi_glenn

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    if this is homework question you should post it there, and also show work done / attempt to solution.
     
  4. Oct 7, 2007 #3

    uae

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    can you help me to solve it plzz my doctor need it tomorow

    how to solve it give me some idia
     
  5. Oct 7, 2007 #4

    malawi_glenn

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    Use the relation that closes approach means that the kinetic energy of the alpha is zero so you only have potential energy; that is given by the Coloumb law, and Angular momentum L = mvr

    d=p\2(1+cosecΘ\2)
    Where p is defined in fig .

    Where is p in fig? p = mv; i.e linear momentum.
     
  6. Oct 8, 2007 #5

    uae

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    Hi

    [​IMG]

    P IN RATHERFORD SCUTTERING IN FIG.

    but p=Zze^2/4(3.14)EokE

    Eo=8.85*10^-12C^2/N-m^2
     
  7. Oct 8, 2007 #6

    malawi_glenn

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    i have no idea what you are talking about now or writing. Can you define all constants that you are using?

    What is Eo? There is no structure in your work now..

    This derivation is done in millions of textbooks in introductory nuclear physics, and also in some books in classical mechanics to (this is the classical Rutherford-scattering-crossection we are discussing now)
     
  8. Oct 8, 2007 #7

    uae

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    Epsilon is εo=8.85*10^-12C^2/N-m^2
     
  9. Oct 8, 2007 #8

    malawi_glenn

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    You have still not structured all the forumulas and made them readble, and also you have not show any work done.

    Perhaps you should try google it, or consult a textbook from you library, in this way you can't get help here.
     
  10. Oct 8, 2007 #9

    uae

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    ok thank you
     
  11. Jan 14, 2011 #10
    Hi,
    I am also wondering about the first question 'uae' posed. Since he did not explain it correctly, I will elaborate a bit. 'p' is the distance of closest approach for b = 0 (b is the impact parameter). At this point, the incident kinetic energy is transformed into potential energy in the coulomb field.
    So I have: 1/2mv^2 = (Zze^2)/(4piEop) where Z is atomic number, z is charge, e is value of electron, Eo is permittivity of vacuum. This gets further reduced to a value of
    p = (Zze^2)/(4piEoT) where T is the kinetic energy.
    Now I get confused. I'm asuming I will have to use conservation of energy, as well as angular momentum, but how? This question is not really proved in any textbooks that I have seen, at least not in the format of how I am to report the answer. Any help would be awesome, thanks!
     
  12. Feb 1, 2011 #11
    can anybody explain rutherford scattering in classical mechanics?
     
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