The species having identical radii according to the Bohr's theory are:

A. H (first orbit)

B. He$$^+$$ (first orbit)

C. He$$^+$$ (Second orbit)

D. Li$$^{2+}$$ (first orbit)

E. Be$$^{3+}$$ (Second orbit)
Choose the correct answer from the options given below: 

For a hydrogen-like species (only one electron), Bohr’s radius formula is

$$r_n = a_0\,\frac{n^{2}}{Z}$$
where $$a_0 = 0.529\ \text{Å}$$ is the Bohr radius, $$n$$ is the principal quantum number (orbit number) and $$Z$$ is the nuclear charge.

Calculate $$r_n$$ for every species mentioned:

Case A: $$\text{H (1st orbit)}$$  ⇒  $$Z = 1,\ n = 1$$
$$r = a_0\,\frac{1^{2}}{1} = a_0$$

Case B: $$\text{He}^+ \text{ (1st orbit)}$$  ⇒  $$Z = 2,\ n = 1$$
$$r = a_0\,\frac{1^{2}}{2} = \frac{a_0}{2}$$

Case C: $$\text{He}^+ \text{ (2nd orbit)}$$  ⇒  $$Z = 2,\ n = 2$$
$$r = a_0\,\frac{2^{2}}{2} = 2a_0$$

Case D: $$\text{Li}^{2+} \text{ (1st orbit)}$$  ⇒  $$Z = 3,\ n = 1$$
$$r = a_0\,\frac{1^{2}}{3} = \frac{a_0}{3}$$

Case E: $$\text{Be}^{3+} \text{ (2nd orbit)}$$  ⇒  $$Z = 4,\ n = 2$$
$$r = a_0\,\frac{2^{2}}{4} = a_0$$

Comparing the radii:

• Case A gives $$r = a_0$$.
• Case E also gives $$r = a_0$$.

No other cases match each other.

Hence the species having identical radii are H (1st orbit) and Be$$^{3+}$$ (2nd orbit).

Option B which is: A, E Only