The radius of fifth orbit of Li$$^{++}$$ is ______ $$\times 10^{-12}$$ m. Take: radius of hydrogen atom = 0.51 $$\mathring{A}$$

The radius of the $$n$$th orbit in a hydrogen-like atom is given by

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

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

For Li$$^{++}$$, we have $$Z = 3$$ and $$n = 5$$, so

$$r_5 = \frac{5^2 \times a_0}{3} = \frac{25 \times 0.51}{3}\;\mathring{A}$$ $$r_5 = \frac{12.75}{3} = 4.25\;\mathring{A} = 4.25 \times 10^{-10}\;\text{m} = 425 \times 10^{-12}\;\text{m}$$

Hence, the radius of the fifth orbit of Li$$^{++}$$ is $$425 \times 10^{-12}$$ m. So, the answer is $$425$$.