Question: If $r_1$ and $r_2$ are the radii of atomic nuclei of mass numbers $64$ and $27$ respectively, then the value of $\left(\frac{r_1}{r_2}\right)$ is:

Step 1 (Formula / Governing Law & Given Data): [0.25 Mark]

The radius $R$ of a nucleus with mass number $A$ is given by the empirical relation:

$$R = R_0 A^{1/3}$$

Where $R_0$ is a constant (approx. $1.2 \times 10^{-15} \text{ m}$).

Given: $A_1 = 64$, $A_2 = 27$.

Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]

Using the formula for both nuclei:

$$r_1 = R_0 (A_1)^{1/3}$$ $$r_2 = R_0 (A_2)^{1/3}$$

Taking the ratio:

$$\frac{r_1}{r_2} = \frac{R_0 (A_1)^{1/3}}{R_0 (A_2)^{1/3}} = \left( \frac{A_1}{A_2} \right)^{1/3}$$

Substituting the given values:

$$\frac{r_1}{r_2} = \left( \frac{64}{27} \right)^{1/3}$$ $$\frac{r_1}{r_2} = \left( \frac{4^3}{3^3} \right)^{1/3} = \frac{4}{3}$$

Step 3 (Final Concluded Answer): [0.25 Mark]

The value of the ratio is 4/3.

Correct Option: (B)

💡 EL10 Concept Insight: Think of a nucleus like a ball of clay; because the density of all nuclei is roughly the same, if you add more "stuff" (mass number), the radius only grows by the cube root of that amount, which is why we use the $1/3$ power.