The radius of the 2$$^{nd}$$ orbit of Li$$^{2+}$$ is $$x$$. The expected radius of the 3$$^{rd}$$ orbit of Be$$^{3+}$$ is

The radius of the nth orbit in a hydrogen-like atom is:

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

where $$a_0$$ is the Bohr radius and $$Z$$ is the atomic number.

For Li²⁺ (Z = 3), 2nd orbit:

$$x = r_2 = \frac{4a_0}{3}$$

For Be³⁺ (Z = 4), 3rd orbit:

$$r_3 = \frac{9a_0}{4}$$

Ratio:

$$\frac{r_3}{x} = \frac{9a_0/4}{4a_0/3} = \frac{9a_0}{4} \times \frac{3}{4a_0} = \frac{27}{16}$$

So $$r_3 = \frac{27}{16}x$$

The correct answer is Option 3: $$\frac{27}{16}x$$.