Let the masses of the two nuclear parts be $$m_1$$ and $$m_2$$, and their respective speeds be $$v_1$$ and $$v_2$$. According to the problem, the ratio of their nuclear sizes is $$1 : 2^{1/3}$$. Since the nuclear size is proportional to the cube root of the mass, we can write:

$$ \frac{m_1}{m_2} = \left(\frac{1}{2^{1/3}}\right)^3 = \frac{1}{2} $$

Now, according to the conservation of linear momentum, the momentum before disintegration is equal to the momentum after disintegration:

$$ m_1 v_1 = m_2 v_2 $$

From the problem statement, the ratio of their respective speeds is $$n : 1$$, so we can write:

$$ v_1 = n \cdot v_2 $$

Substitute the expression for $$v_1$$ into the momentum conservation equation:

$$ m_1 (n \cdot v_2) = m_2 v_2 $$

We know the mass ratio, so substitute that into the equation:

$$ \frac{1}{2} m_2 (n \cdot v_2) = m_2 v_2 $$

Divide both sides by $$m_2 v_2$$:

$$ \frac{1}{2} n = 1 $$

Now, solve for $$n$$:

$$ n = 2 $$

Thus, the value of $$n$$ is 2.