Concept:
Experimental observations demonstrate that nuclear matter has a nearly constant density. As a result, the volume of an atomic nucleus is directly proportional to the total number of nucleons (protons + neutrons) it contains, which is defined as its mass number (\(A\)). Assuming a spherical nuclear geometry, this relationship is given by:
\[
\text{Volume} = \frac{4}{3}\pi r^3 \propto A \quad \Rightarrow \quad r \propto A^{1/3}
\]
This allows us to write the nuclear radius formula as:
\[
r = r_0 A^{1/3}
\]
where \(r_0\) is an empirical constant representing the approximate radius of a single nucleon (\(r_0 \approx 1.2 \times 10^{-15}\text{ m} = 1.2\text{ fm}\)).
Step 1: Setting up the ratio equation based on given mass numbers.
We are given two nuclei with different mass numbers:
• Mass number of the first nucleus: \(A_1 = 64\)
• Mass number of the second nucleus: \(A_2 = 27\)
Using the radius proportional formula for both nuclei:
\[
r_1 = r_0 (A_1)^{1/3}
\]
\[
r_2 = r_0 (A_2)^{1/3}
\]
Dividing the equation for \(r_1\) by the equation for \(r_2\):
\[
\frac{r_1}{r_2} = \frac{r_0 (A_1)^{1/3}}{r_0 (A_2)^{1/3}} = \left( \frac{A_1}{A_2} \right)^{1/3}
\]
Step 2: Calculating the final numerical value.
Substitute the mass numbers \(A_1 = 64\) and \(A_2 = 27\) into the ratio expression:
\[
\frac{r_1}{r_2} = \left( \frac{64}{27} \right)^{1/3}
\]
Recognizing that both numbers are perfect cubes:
\[
64 = 4^3 \quad \text{and} \quad 27 = 3^3
\]
We can rewrite the expression as:
\[
\frac{r_1}{r_2} = \left( \frac{4^3}{3^3} \right)^{1/3} = \left[ \left(\frac{4}{3}\right)^3 \right]^{1/3} = \frac{4}{3}
\]
Thus, the value of the ratio \(\frac{r_1}{r_2}\) is equal to \(\frac{4}{3}\), which matches Option (B).