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Nuclear Physics - mean-square charge radius of a uniformly charged sphere

  1. Feb 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the mean-square charge radius of a uniformly charged sphere (with radius R) is
    < r^2 > = 3*R^2 / 5

    2. Relevant equations

    < r^2 > = [tex]\int[/tex] [tex]\varphi[/tex]f* r^2 [tex]\varphi[/tex]i dV

    [tex]\varphi[/tex]f* = exp(-i q dot r)

    [tex]\varphi[/tex]i = exp(i q dot r)

    where q = kf - ki
    and pf = h_bar * kf
    pi = h_bar * ki

    r > R: V(r) = -Ze^2 / (4*pi*e0) * 1/r^2
    r < R: V´(r) = -Ze^2 / (4*pi*e0*R) * (3/2 - 1/2 * (r/R)^2)

    Ei = 1/2*m*vi

    Er0 = J0^2/(2*m*r0^2) + V(r0) or V´(r0) according as r > R or r < R.

    J0 = Ji = m*vi*b

    where b is the impact parameter and is b = Ze^2/ (4*pi*e0*m*vi^2) * cot(A/2) for hyperbolic orbits, Ei > 0

    3. The attempt at a solution

    I basically want to know if my approach is correct. I think that I have to find < r0^2 > where r0 is the distance of closest approach to the centre of the charge distribution either inside it (where the electron feels V´(r)) or outside it (where the electron feels V(r)).

    I've done some calculations using the radial energy equation, trying to solve for r0^2, but the calculations are a mess and I get a feeling I'm going about it all wrong. The problem is from Krane's Introductory Nuclear Physics, Chapter 3, problem 3.1.
     
  2. jcsd
  3. Feb 19, 2009 #2

    malawi_glenn

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    Homework Helper

    You are mixing concepts here, you have tried to evaluate the Form factor, F, with the potential r^2 ..

    you want to evaluate the mean square radius, then use formula for distributions:
    [tex]1 = A\int \rho (\vec{r})d\vec{r}[/tex]

    this is from mathematical statistics. Now use that the distribution of charge is constant up radius R, and for radius larger than R it is zero. Angular integration gives you 4pi, so it is easy to find the normalization constant A.

    Then you recall from statistics that the average value of some quantity Q is
    [tex]<Q> = A\int Q\rho (\vec{r})d\vec{r}[/tex]

    Now you have what you need.
    What you did wrong was to mix this up with the Form factor F.
     
  4. Feb 19, 2009 #3
    Hi Malawi Glenn,

    Thank you very much, I used the equations you suggested and got the required result.

    I'm concurrently registered for Stat Mech, QM, Nuclear Physics and Solid State Phys this year and realised when I attempted this problem that I should have completed Stat Mech and QM before I took Nuc Phys and SS Phys, but what's done is done and I'll just have to cope.

    Thanks again.
     
  5. Feb 19, 2009 #4

    malawi_glenn

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    well yes, QM is pre requirement for both SS phys and Nuc phys. But the math behind is just probability, continuous and discrete.
     
  6. Oct 9, 2011 #5
    I know it is too late but I have a question.

    How could we find the normalization constant?

    We have: ρ=3/(4.pi.R^3).

    I know that we should also have A=4.pi, but the integration:

    1= A ∫ρ dr

    does not give that result. Any ideas?
     
  7. Oct 11, 2011 #6
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