1. Jan 31, 2008

### 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?

Last edited: Jan 31, 2008
2. Jan 31, 2008

### pam

The mean radius is not generally used.

3. Jan 31, 2008

### russdot

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

4. Jan 31, 2008

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

Last edited: Jan 31, 2008