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

[russdot]

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 $$\int$$p(r)*r*p(r) dV ?
Which would make <r$$^{2}$$>$$^{1/2}$$ = ($$\int$$p(r)*r$$^{2}$$*p(r) dV)$$^{1/2}$$, correct?

 

[pam]

The mean radius is not generally used.
Your equation for the rms radius is correct.

 

[russdot]

Great, thanks!
I'm assuming if the rms equation is correct, then the mean value equation is also correct..

 

[malawi_glenn]

I have always used:

$$<r^2> = \int \rho (r)r^2 d\vec{r}$$

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:
$$<r^2> = \int \rho (r)r^2 d\vec{r} / \int \rho (r) d\vec{r}$$