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Mean radius, r.m.s. radius of nucleus

  1. Jan 31, 2008 #1
    Hello,
    Given a particular charge distribution p(r) = p_0*exp(-r[tex]^{2}[/tex]/a[tex]^{2}[/tex]), I was wondering if the proper way to calculate the mean radius <r> would be [tex]\int[/tex]p(r)*r*p(r) dV ?
    Which would make <r[tex]^{2}[/tex]>[tex]^{1/2}[/tex] = ([tex]\int[/tex]p(r)*r[tex]^{2}[/tex]*p(r) dV)[tex]^{1/2}[/tex], correct?
     
    Last edited: Jan 31, 2008
  2. jcsd
  3. Jan 31, 2008 #2

    pam

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    The mean radius is not generally used.
    Your equation for the rms radius is correct.
     
  4. Jan 31, 2008 #3
    Great, thanks!
    I'm assuming if the rms equation is correct, then the mean value equation is also correct..
     
  5. Jan 31, 2008 #4

    malawi_glenn

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    I have always used:

    [tex] <r^2> = \int \rho (r)r^2 d\vec{r} [/tex]

    since the wave function(s): [tex] \psi (r)^* \cdot \psi (r) = \rho (r) [/tex]

    If the density is normalised to unity: [tex] \int \rho (r) d\vec{r} = 1 [/tex]

    Otherwise:
    [tex] <r^2> = \int \rho (r)r^2 d\vec{r} / \int \rho (r) d\vec{r} [/tex]
     
    Last edited: Jan 31, 2008
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