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Radius of curvature R of alpha particle beam

  1. May 5, 2014 #1
    1. The problem statement, all variables and given/known data

    test.jpg

    2. Relevant equations
    r=mv/qb
    mv=sqrt(2*KE*m(alpha))
    m(alpha)=6.64e-27 kg

    3. The attempt at a solution
    i was just wondering how to get the answer (7.6e-4 m). i get path curving down, and do
    r=sqrt(2*1e3eV*6.64e-27kg*1.6e-19J/eV)/(q*B)
    =15.18e-4m

    so to get the right answer i would have to divide this by 2..why is this?? i used the right equation but am I doing something wrong?
     
  2. jcsd
  3. May 5, 2014 #2

    TSny

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    What did you use for the charge of the alpha particle?
     
  4. May 5, 2014 #3
    charge of alpha particle is +2. so if its only alpha particle I have to divide by 2? if it was electron (-1) I would not need to divide by anything??
     
  5. May 5, 2014 #4

    TSny

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    You didn't show what number you used for q. It should be in Coulombs. What is the charge in Coulombs of an alpha particle?
     
  6. May 5, 2014 #5
    I'm guessing its +2 times charge of electron. q=1.6e-19C
     
  7. May 6, 2014 #6

    TSny

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    Right. An alpha particle has a charge q of twice the magnitude of the charge of an electron.

    But, the magnitude of the charge on an electron is 1.6 x 10-19 C.

    So, if q is 2 times the magnitude of the charge on an electron, then q ≠ 1.6 x 10-19 C

    What value did you use for q when you got the wrong result of 15.18 x 10-4 m?
     
  8. May 6, 2014 #7
    I was using q=1.6e-19C for electron. I guess it should've been 3.2e-19C but I forgot to multiply by the +2 charge. thanks.
     
  9. May 6, 2014 #8

    TSny

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    Good. When using r = mv/qB, q is the charge of the particle that is moving in the magnetic field.
     
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