A nucleus has mass number $$\alpha$$ and radius $$R_{\alpha}$$ Another nucleus has mass number $$\beta$$ and radius $$R_{\beta}$$. If $$\beta=8\alpha$$ then $$R_{\alpha}/R_{\beta}$$ is:

A nucleus with mass number $$\alpha$$ has radius $$R_\alpha$$ and another with mass number $$\beta$$ has radius $$R_\beta$$. Given $$\beta = 8\alpha$$, we apply the nuclear radius formula $$R = R_0 A^{1/3}$$, where $$A$$ is the mass number and $$R_0$$ is a constant.

Substituting for the two nuclei yields $$\frac{R_\alpha}{R_\beta} = \frac{R_0 \alpha^{1/3}}{R_0 \beta^{1/3}} = \left(\frac{\alpha}{\beta}\right)^{1/3}$$. Using $$\beta = 8\alpha$$ gives $$\left(\frac{\alpha}{8\alpha}\right)^{1/3} = \left(\frac{1}{8}\right)^{1/3} = \frac{1}{2} = 0.5$$.

The correct answer is Option 3: 0.5.