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Probing Atomic Nuclei with Electron Scattering.

  1. Feb 13, 2007 #1
    1. The problem statement, all variables and given/known data
    To study the structure of the lead nucleus, electrons are fired at a lead target. Some of the electrons actually enter the nuclei of the target, and the deflection of these electrons is measured. The deflection is caused by the charge of the nucleus, which is distributed approximately uniformly over the spherical volume of the nucleus. A lead nucleus has a charge of + 82e and a radius of R = 7.10×10^−15

    Find the acceleration of an electron at a distance of 3.00 R from the center of a lead nucleus.

    2. Relevant equations

    Charge of Electron = -1.60*10^-19

    Mass of Electron = 9.11*10^-31

    [tex] F = \frac {k*q*Q} {r^{2}} [/tex]

    [tex] k = \frac {1} {4*pi*e_o} [/tex]

    [tex] e_o = 8.85*10^{-12} [/tex]

    [tex] A = \frac {F} {m} [/tex]

    3. The attempt at a solution

    => [tex] F = \frac {\frac {1} {4*pi*8.85*10^{-12}}*-1.60*10^{-19}*82*1.60*10^{-19}} {(3*(7.10*10^{-15})^{2}} [/tex]
    Last edited: Feb 13, 2007
  2. jcsd
  3. Feb 13, 2007 #2


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    Staff: Mentor

    Looks good so far.
  4. Feb 13, 2007 #3
    When I get down to a numeric answer, it doesn't seem right.

    [tex] \frac {(6.95*10^{-12}*(-2.17*10^{-36}} {1.51*10^{-28}} [/tex]

    = [tex] -9.9*10^{-20} [/tex]
  5. Feb 13, 2007 #4


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    Staff: Mentor

    First, remember that you are solving for a=F/m. Second, the first term in the numerator went from 1/10^-12 to 10^-12. Maybe a misplaced "-" sign?
  6. Feb 13, 2007 #5
    Yes you're right the first term was actually incorrect:

    [tex] \frac {(8.99*10^{9})*(-2.17*10^{-36})} {1.51*10^{-28}} [/tex]

    F = 1.00063

    [tex] A = \frac {F} {M} [/tex]

    [tex] A = \frac {1.00063} {9.11*10^{-31}} [/tex]

    [tex] A = 1.09*10^{30} [/tex]

    Can you confirm this numeric answer? I think it should be 10^32?

    I solved for the case when it's just distance of "R" and it came out to be 10^32, so shouldn't these other distances have a factor of the same 10^32? I have limited of tries and I just wanted to be sure I wasn't doing the math incorrectly and making a careless mistake. Thank you
  7. Feb 13, 2007 #6


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    Staff: Mentor

    I get a different denominator number for F, but I could just be messing up the calculator (I had to borrow one). I have to bail for a while -- just re-check your numbers. One trick is to punch the numbers into the calculator in a different order to help you catch any calculation errors. Good luck.
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