What Determines the r_0 Constant in Nuclear Radius Calculations?

[mateomy]

I've been noticing that for expressions finding the nuclear radius there is a constant term $r_0$, that I can't seem to find an explanation for. The full term is
$$R=r_{0}A^{1/3}$$
That the atomic radius is roughly the 3rd root of the number of nucleons. But where is the $r_0$ coming from? I'm using Cottingham and Wong as sources but both of them fail to mention it's derivation. Plus, they both give ${slightly}$ different values, without looking I think Wong assigns it a value of 1.7 fm and Cottingham 1.1 fm. Can anyone explain this to me?


Thanks.

 

[TSny]

Nuclei are sort of fuzzy and don't have definite radii (they're not even necessarily spherical). The formula gives an approximate value of the radius and the value of r_o in the formula is determined from various experiments. See http://en.wikipedia.org/wiki/Nuclear_size.

 

[tia89]

This is I think an expression found by experimental or at least phenomenological means. As there $A$ is the number of nucleons, you would have that $r_0$ is the size of a single nucleon (and has to be determined experimentally of course). It is needed in this expression formally to give the right dimension to $R$.

Also notice that (something I think is interesting) if you compute the volume of the nucleus (interpreting really $r_0$ as the radius of the single nucleon), you will have that the total volume is $A$ times the volume of a nucleon, which is clearly consistent with what one would imagine.

 

[tia89]

the formula is determined from various experiments. See http://en.wikipedia.org/wiki/Nuclear_size.

Looking in the Wikipedia article to the values given for the experimental sizes of neutron and proton, indeed $r_0$ is a constant determined empirically and seems to me a sort of mean value of the two.