1. Use Bohr’s radius formula for hydrogen-like species

For a one-electron ion, the radius of the $n^{\text{th}}$ orbit is

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

where:

• $a_0$ = Bohr radius
• $n$ = orbit number
• $Z$ = atomic number

2. Radius of the 2nd orbit of $\mathrm{Li^{2+}}$

$\mathrm{Li^{2+}}$ is a hydrogen-like ion with $Z=3$.

So,

$x = r_2(\mathrm{Li^{2+}}) = a_0\frac{2^2}{3} = a_0\frac{4}{3}$

3. Radius of the 3rd orbit of $\mathrm{Be^{3+}}$

$\mathrm{Be^{3+}}$ is also hydrogen-like with $Z=4$.

Thus,

$r_3(\mathrm{Be^{3+}}) = a_0\frac{3^2}{4} = a_0\frac{9}{4}$

4. Express this in terms of $x$

Since

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

we get

$r_3(\mathrm{Be^{3+}})=a_0\frac{9}{4} = \left(a_0\frac{4}{3}\right)\cdot \frac{9/4}{4/3}$

So,

$r_3(\mathrm{Be^{3+}})=x\cdot \frac{9}{4}\cdot \frac{3}{4} = x\cdot \frac{27}{16}$

Hence,

$r_3(\mathrm{Be^{3+}})=\frac{27}{16}x$

5. Option check

• A: $\frac{16}{27}x$ ❌
• B: $\frac{4}{9}x$ ❌
• C: $\frac{9}{4}x$ ❌
• D: $\frac{27}{16}x$ ✅

Therefore, the correct answer is D.