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

  • Thread starter russdot
  • Start date
  • #1
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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:

Answers and Replies

  • #2
pam
455
1
The mean radius is not generally used.
Your equation for the rms radius is correct.
 
  • #3
16
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Great, thanks!
I'm assuming if the rms equation is correct, then the mean value equation is also correct..
 
  • #4
malawi_glenn
Science Advisor
Homework Helper
4,786
22
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]
 
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