A nucleus has mass number $$A_1$$ and volume $$V_1$$. Another nucleus has mass number $$A_2$$ and volume $$V_2$$. If relation between mass number is $$A_2 = 4A_1$$, then $$\frac{V_2}{V_1} =$$

We need to find the ratio of volumes $$V_2/V_1$$ of two nuclei with mass numbers $$A_1$$ and $$A_2 = 4A_1$$.

The radius of a nucleus is given by the empirical relation $$R = R_0 \, A^{1/3}$$, where $$R_0 \approx 1.2 \, \text{fm}$$ is a constant and $$A$$ is the mass number. Since the nucleus is approximately spherical, its volume is $$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3 \, A$$, which shows that the nuclear volume is directly proportional to the mass number: $$V \propto A$$.

Thus the volumes of the two nuclei can be expressed as $$V_1 = \frac{4}{3}\pi R_0^3 \, A_1$$ and $$V_2 = \frac{4}{3}\pi R_0^3 \, A_2$$. Taking the ratio gives $$\frac{V_2}{V_1} = \frac{A_2}{A_1}$$, and substituting $$A_2 = 4A_1$$ leads to $$\frac{V_2}{V_1} = \frac{4A_1}{A_1} = 4$$.

The answer is 4.