Discussion
According to the scattering experiments of electrons from various nuclei, the nuclear charge density \(\rho_{ch}\) as a function of radial distance r is approximated by relation .............
(Where \(\rho_{0}\) is constant nucleon density in the core, a is the diffusiveness parameter and R is the average radius of particular nucleus.)
Answer: A
The charge density is \(\rho_{ch}(r) = \frac{\rho_{0}}{1 + \exp\left(\frac{r - R}{a}\right)}\).
This is the Fermi (Woods-Saxon) distribution fitted to electron-scattering data. Check its shape:
- deep inside (\(r \ll R\)) the exponential is tiny, so \(\rho \approx \rho_{0}\), a constant core density;
- at \(r = R\) the density is \(\rho_{0}/2\), so \(R\) is the half-density radius;
- far outside (\(r \gg R\)) the exponential grows and \(\rho\) falls smoothly to zero over a skin set by \(a\).
With \(r + R\) in the exponent the density would never reach \(\rho_{0}\) and would have no surface at \(R\). With \(1 - \exp\) in the denominator it would blow up at \(r = R\) and turn negative outside, which a density cannot do.
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