Considering the Bohr model of hydrogen like atoms, the ratio of the radius $$5^{\text{th}}$$ orbit of the electron in $$Li^{2+}$$ and $$He^+$$ is

For a hydrogen-like ion, Bohr’s radius of the electron in the $$n^{\text{th}}$$ orbit is given by

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

where $$a_0$$ is the Bohr radius of the first orbit of the hydrogen atom and $$Z$$ is the atomic number of the nucleus.

Both $$Li^{2+}$$ and $$He^{+}$$ have a single electron, so the same formula applies.

Case 1: $$Li^{2+}$$
For lithium, $$Z_{Li}=3$$ and for the $$5^{\text{th}}$$ orbit, $$n=5$$.

$$r_{Li} = \frac{(5)^2\,a_0}{3}= \frac{25a_0}{3}$$

Case 2: $$He^{+}$$
For helium, $$Z_{He}=2$$ and again $$n=5$$.

$$r_{He} = \frac{(5)^2\,a_0}{2}= \frac{25a_0}{2}$$

Now compute the required ratio:

$$\frac{r_{Li}}{r_{He}} = \frac{\dfrac{25a_0}{3}}{\dfrac{25a_0}{2}} = \frac{2}{3}$$

Hence, the ratio of the radii is $$\dfrac{2}{3}$$.

Option D is correct.