(a) Assuming nuclei are spherical in shape, show that its radius \((r)\) is proportional to the cube root of mass number \((A) .(\mathrm{b})\) In general, the radius of a nucleus is given by \(r=r_{0} A^{\frac{1}{3}},\) where \(r_{0},\) the proportionality constant, is given by \(1.2 \times 10^{-15} \mathrm{~m}\). Calculate the volume of the \({ }^{238} \mathrm{U}\) nucleus.

One can show the proportional relation between the radius \(r\) and the cube root of the mass number \(A\) through dimensional analysis and the knowledge of the geometry of spheres. For a sphere of mass \(M\) and radius \(r\), the volume \(V\) is given by \(V = \frac{4}{3}\pi r^3\). As nuclei are assumed to be spheres of uniform density \(\rho\) so \(M = \rho V = \rho (\frac{4}{3}\pi r^3)\). But the mass of a nucleus is also proportional to its mass number \(A\), denoted as \(M = kA\) where \(k\) is a constant of proportionality. By equating these two expressions for \(M\), \(\rho (\frac{4}{3}\pi r^3) = kA\), the proportionality between \(r\) and the cube root of \(A\) can be derived: \(r \propto A^{\frac{1}{3}}\).

Now, the aim is to find the volume of a uranium-238 nucleus. The radius \(r\) of this nucleus can be calculated by substituting \(A = 238\) and \(r_0 = 1.2 \times 10^{-15} m\) into the given formula \(r = r_{0}A^{\frac{1}{3}}\).

With the radius calculated in the previous step,- the volume of the uranium-238 nucleus can be found using the formula for the volume of a sphere: \(V = \frac{4}{3}\pi r^3\).