Natural Units and the Nuclear Radius

Question

In a paper I'm looking at and also in a similar program, the authors note that the nuclear radius $R$ is given in "natural units" as $R = 3.11\times 10^{-3}\;A^{1/3}$. This has me slightly confused, as I seem to be missing a factor of 2. My understanding is that natural units usually entail $\hbar = c = 1$, so that length is measured in units of inverse energy. A typical estimate of the nuclear radius as a function of nucleon number $A$ is $R = 1.2\, \text{fm}$, so (I thought) $R_{natural} = R/\hbar c$ and thus $$ R = \frac{1.2\,A^{1/3}\,\text{fm}}{197.327 \, \text{MeV-fm}} = 6.1 \times 10^{-3} \, A^{1/3} $$ in natural units, which is about twice what I expect to get. I'm not overly comfortable in these units yet, so I can't spot the error. Where is this factor of two coming from?

Answer 1

I think the answer is found in another reference on beta decay, similar to where this question originated. It describes the nuclear radius R as being expressed in terms of the electron Compton wavelength. Then $$ R_{natural} = \frac{R}{\hbar\,/\,m_ec} = R\frac{m_ec^2}{\hbar c} \approx \frac{(1.2\,\text{fm})\, (0.511 \,\text{MeV})}{197.327\,\text{MeV fm}}A^{1/3} \approx 3.3 \times 10^{-3} A^{1/3}, $$ if I make the assumption the authors mean the reduced Compton wavelength and again set $\hbar = c = 1$.

This makes sense in the context of beta decay (which I forgot to mention I was looking at originally). Nuclear physics is normally non-relativistic, but beta decay results in relativistic electrons and neutrinos and we must handle them relativistically. Integrals are then often made dimensionless by expressing energies in the ratios $\epsilon = E\,/\,m_e$, so a dimensionless nuclear radius is logical here.