Identical Radii Species in Bohrs Theory

To find the species having identical radii according to Bohr's theory, we use the formula for the radius of the $n$-th orbit of a hydrogen-like ion with atomic number $Z$:

$r_n = r_0 \cdot \frac{n^2}{Z}$

where $r_0 = 0.529 \text{ \AA}$ is the Bohr radius for a hydrogen atom in its ground state.

For two species to have identical radii, their value of the factor $\frac{n^2}{Z}$ must be equal. Let's calculate $\frac{n^2}{Z}$ for each given species:

1. A. $\text{H}$ (first orbit):

   • Atomic number $Z = 1$, Principal quantum number $n = 1$
   • $\frac{n^2}{Z} = \frac{1^2}{1} = 1$
   • Radius $r_A = 1 \cdot r_0 = 0.529 \text{ \AA}$

2. B. $\text{He}^+$ (first orbit):

   • Atomic number $Z = 2$, Principal quantum number $n = 1$
   • $\frac{n^2}{Z} = \frac{1^2}{2} = 0.5$
   • Radius $r_B = 0.5 \cdot r_0 = 0.2645 \text{ \AA}$

3. C. $\text{He}^+$ (second orbit):

   • Atomic number $Z = 2$, Principal quantum number $n = 2$
   • $\frac{n^2}{Z} = \frac{2^2}{2} = 2$
   • Radius $r_C = 2 \cdot r_0 = 1.058 \text{ \AA}$

4. D. $\text{Li}^{2+}$ (first orbit):

   • Atomic number $Z = 3$, Principal quantum number $n = 1$
   • $\frac{n^2}{Z} = \frac{1^2}{3} = \frac{1}{3}$
   • Radius $r_D = \frac{1}{3} \cdot r_0 \approx 0.1763 \text{ \AA}$

5. E. $\text{Be}^{3+}$ (second orbit):

   • Atomic number $Z = 4$, Principal quantum number $n = 2$
   • $\frac{n^2}{Z} = \frac{2^2}{4} = 1$
   • Radius $r_E = 1 \cdot r_0 = 0.529 \text{ \AA}$

Comparing the computed values, species A ($\text{H}$, first orbit) and species E ($\text{Be}^{3+}$, second orbit) both have $\frac{n^2}{Z} = 1$ and therefore have identical radii.

Hence, the correct option is B (A and E Only).