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Finding normalization constant for proton charge distributio

  1. Sep 29, 2016 #1
    1. The problem statement, all variables and given/known data


    For a model of a proton's charge distribution, hw1eqn2.gif :


    • I have to find the constant of normalisation for rho.

    2. Relevant equations


    3. The attempt at a solution

    I wrote p(r)=p_0 * (e^-r/R)/r

    I then wrote it as
    p_0^2 * integral from -infinity to +infinity of (e^-(r/)^2)/r^2 = 1

    i tried to change variable by making z = r/R but then i was left with

    1/R*p_0^2* integral of e^(-z^2) * (z^-2) = 1

    I know that the integral of e^(-z^2) is the square root of pi but I dont know how to evaluate this with the z^-2 as well.

    Have I started this correctly at all?

    Thanks
     
  2. jcsd
  3. Sep 29, 2016 #2

    blue_leaf77

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    You made mistake when calculating ##(e^{-r/R})^2##.
     
  4. Sep 29, 2016 #3

    TSny

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    What does this integral represent? Why are you squaring the charge density? Why should the integral equal 1?
     
  5. Sep 30, 2016 #4
    Part of my problem I think is that I'm not 100% sure on why I'm doing it. In my quantum work before the wave function was squared and multiplied by its complex conjugate and summed over infinity because the probability of finding it there will be 1. (I think). So I've just tried to use the same kind of method for this problem. Thanks
     
  6. Sep 30, 2016 #5

    blue_leaf77

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    You are right, I completely misunderstood that radial function to be a wavefunction, didn't pay attention that it's a charge distribution already.
     
  7. Sep 30, 2016 #6
    So should I not have squared this at all?
     
  8. Sep 30, 2016 #7

    blue_leaf77

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    What have you learned from your electrodynamics course about charge distribution? For example given total charge ##Q##, its density ##\rho(\mathbf r)##, and volume ##V##, what's the equation connecting these quantities?
     
  9. Sep 30, 2016 #8

    kuruman

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    Your attempt at a solution is incomplete. What did you use for dV in your integral?
     
  10. Sep 30, 2016 #9
    oh yeah

    rho = total charge / volume right or (ze)/(4/3pi*R^3) ?

    and rho in this problem also = constant * (e^(r/R))/r

    I think I need a volume integral here somewhere but Ive not quite reached setting it up yet
     
  11. Sep 30, 2016 #10

    kuruman

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    This is true only if the volume charge density is constant, i.e. independent of r. Here it is not because it is given to you as a function of r.
    You do need a volume integral that looks like ## \int \rho (r) dV ##. You know what ## \rho (r) ## looks like. What about ## dV ##?
     
  12. Sep 30, 2016 #11
    or maybe that means that i should integrate
    dV would look like r^2 sin theta right? if i do this then doesn't that mean that the integral is equal to the total charge?
     
  13. Sep 30, 2016 #12

    kuruman

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    Not right. A volume element dV in spherical coordinates is more than just r2sinθ. What is the correct form?
    Only if you integrate over the correct limits of integration. What are these limits?
     
  14. Sep 30, 2016 #13
    r^2 sin theta dr dtheta dphi

    limits for r theta and phi are 0 to infinity, 0 to pi, and 0 to 2pi, respectively. but why do I need to find total charge to find this constant of normalisation?

    thanks for your patience and advice
     
  15. Sep 30, 2016 #14

    kuruman

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    The proton has charge e = 1.6×10-19 C. The integral ## \int \rho (r) dV ## with the charge density you have will not give you that value or units. You need to multiply the charge distribution by normalization constant N so that the integral indeed gives you that value. This is what normalization means. When you have your answer, you should do a dimensional analysis to verify that your normalized ## \rho (r) ## has dimensions of charge/volume.

    I am not sure that the upper integration limit for r extends all the way to infinity. What is the model for the proton that you are using according to the statement of the problem? Is the proton modeled as extending all the way out to infinity?
     
  16. Sep 30, 2016 #15
    at the bottom of the sheet it has a note saying that the charge distribution falls away as r approaches infinity and that this form is not very realistic but was chosen to simplify the maths so i assumed that meant that the limits were 0 and infinity. does that sound possible?
     
  17. Sep 30, 2016 #16

    kuruman

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    Yes, that's what it meant. Do you understand how to proceed from here?
     
  18. Sep 30, 2016 #17
    from there had p(r)=N*(e^-r/R)/r so had the integral of
    N*(e^-r/R)/r * r^2 sin theta dr dtheta dphi = 1 and evaluated with limits 0>infinity,0>pi,0>2pi to get N = 1/(4*pi*R^2)
     
  19. Sep 30, 2016 #18

    kuruman

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    Why is the integral equal to 1? You are not normalizing a probability which is equal to 1 when you add all probabilities over all space. You want to say that when you add up all the little charge elements ρdV over all space you get the total charge of the proton. So what should the other side of the equation be?
     
  20. Sep 30, 2016 #19
    so is it just equal to z*e or Q? the total charge?
     
  21. Sep 30, 2016 #20

    kuruman

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    What's z for a proton?
     
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