Nuclear Size
It is evident from Rutherford's α-particle scattering experiment that nucleus has a finite size. The results of α-scattering experiment can be explained on the basis of Coulomb's interaction between the α-particle and the nucleus. The distance of closest approach of the α-particle helped in estimating the size of the nucleus. Thus radius of the nucleus is found to be order of 10-14 m.
Another method to study the size is to study the charge distribution of the nucleus by using high energy electrons. This method involves study of elastic scattering of electrons (i.e. of different energies) from target nuclei. The size of nucleus is calculated from this study. There are reasons to assume that nucleus is nearly spherical in shape.
Mathematically
Various scattering experiments show that the nuclear size vary smoothly with mass number as shown in Fig. (1.4).
The size of the nucleus has been measured by studying variety of experiments involving scattering of neutrons, protons, electrons etc. It is found that the volume of the nucleus is directly proportional to the mass number (number of nucleons) of the nucleus.
If \(R\) is the radius of the nucleus of mass number \(A\), then volume of nucleus \(\propto A\)
\[\frac{4}{3}\pi R^3 \propto A\]
or
\[R^3 \propto A\]
or
\[R \propto A^{1/3}\]
or
\[R = R_0 A^{1/3}\]
(1.5)
where \(R_0 = 1.2 \times 10^{-15}\) m is in the range of nuclear force. It is also known as nuclear unit radius. It is believed to be average nuclear size. The value of \(R_0\) is not definite. The electron scattering experiments predict slightly smaller value of \(R_0\) than predicted by neutron scattering experiments. The value of \(R\) varies from 1.2 to 1.6 fermi where 1 fermi = \(10^{-15}\) m. The different values of nuclear radius from these experiments shows that matter and charge are not distributed in similar manner inside the nucleus.
