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

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To calculate the mean radius <r> for a charge distribution p(r) = p_0*exp(-r^{2}/a^{2}), the correct approach involves integrating p(r)*r*p(r) dV. The equation for the root mean square (rms) radius is confirmed as <r^{2}>^{1/2} = (\int p(r)*r^{2}*p(r) dV)^{1/2}. The mean value equation is also validated, with <r^2> expressed as <r^2> = \int \rho (r)r^2 d\vec{r}, where \rho (r) is derived from the wave function. If the density is normalized to unity, the mean radius formula adjusts accordingly. This discussion clarifies the proper calculations for mean and rms radii in nuclear physics.
russdot
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Hello,
Given a particular charge distribution p(r) = p_0*exp(-r^{2}/a^{2}), I was wondering if the proper way to calculate the mean radius <r> would be \intp(r)*r*p(r) dV ?
Which would make <r^{2}>^{1/2} = (\intp(r)*r^{2}*p(r) dV)^{1/2}, correct?
 
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The mean radius is not generally used.
Your equation for the rms radius is correct.
 
Great, thanks!
I'm assuming if the rms equation is correct, then the mean value equation is also correct..
 
I have always used:

&lt;r^2&gt; = \int \rho (r)r^2 d\vec{r}

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

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

Otherwise:
&lt;r^2&gt; = \int \rho (r)r^2 d\vec{r} / \int \rho (r) d\vec{r}
 
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