The radius of a nucleus of mass number $$64$$ is $$4.8$$ fermi. Then the mass number of another nucleus having radius of $$4$$ fermi is $$\frac{1000}{x}$$, where $$x$$ is _________.

Radius of nucleus with $$A_1 = 64$$ is $$R_1 = 4.8$$ fermi. Another nucleus has $$R_2 = 4$$ fermi. Find $$x$$ where $$A_2 = 1000/x$$.

The radius of a nucleus is proportional to the cube root of its mass number:

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

where $$R_0 \approx 1.2$$ fermi. This relationship exists because nucleons are approximately incompressible, so nuclear volume is proportional to the number of nucleons: $$V \propto A$$, and since $$V = \frac{4}{3}\pi R^3$$, we get $$R \propto A^{1/3}$$.

Taking the ratio of the two radii gives

$$ \frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} $$

Substituting the values yields

$$ \frac{4.8}{4} = \left(\frac{64}{A_2}\right)^{1/3}, $$

so

$$ 1.2 = \left(\frac{64}{A_2}\right)^{1/3}. $$

Upon cubing both sides,

$$ (1.2)^3 = \frac{64}{A_2}. $$

Since $$1.2^3 = 1.2 \times 1.2 \times 1.2 = 1.44 \times 1.2 = 1.728$$, it follows that

$$ A_2 = \frac{64}{1.728}. $$

Noting that $$1.728 = \frac{1728}{1000}$$ and $$1728 = 12^3$$ while $$64 = 4^3$$, we find

$$ A_2 = \frac{64 \times 1000}{1728} = \frac{64000}{1728} = \frac{1000}{27} $$

(since $$64000/1728 = 64/1.728 = 1000/27$$).

Since $$A_2 = \frac{1000}{x}$$, this gives $$x = 27$$.

The answer is 27.