The radius of 2$$^{\text{nd}}$$ orbit of He$$^+$$ of Bohr's model is $$r_1$$ and that of fourth orbit of Be$$^{3+}$$ is represented as $$r_2$$. Now the ratio $$\frac{r_2}{r_1}$$ is $$x : 1$$. The value of $$x$$ is _____.

Find the ratio $$\frac{r_2}{r_1}$$ where $$r_1$$ is the radius of the 2nd orbit of $$He^+$$ and $$r_2$$ is the radius of the 4th orbit of $$Be^{3+}$$.

$$r_n = a_0 \cdot \frac{n^2}{Z}$$

where $$a_0$$ is the Bohr radius, $$n$$ is the principal quantum number, and $$Z$$ is the atomic number.

$$r_1 = a_0 \cdot \frac{2^2}{2} = a_0 \cdot \frac{4}{2} = 2a_0$$

$$r_2 = a_0 \cdot \frac{4^2}{4} = a_0 \cdot \frac{16}{4} = 4a_0$$

$$\frac{r_2}{r_1} = \frac{4a_0}{2a_0} = 2$$

So $$x : 1 = 2 : 1$$, giving $$x = 2$$.

The answer is $$\boxed{2}$$.