In the Bohr model, the radius of the $n^\text{th}$ orbit of a hydrogen-LIKE atom of atomic number $Z$ is $r_n = a_0 n^2 / Z$, where $a_0 = 5.3\times 10^{-11}\,\text{m}$ is the Bohr radius. The radius of the $n = 2$ orbit of $\mathrm{He}^+$ ($Z = 2$) is therefore

1. Bohr's formula for the radius of the $n^\text{th}$ orbit of a hydrogenic atom: $r_n = a_0\,n^2/Z$. 2. For $\mathrm{He}^+$, $Z = 2$. We want $n = 2$. 3. Substitute: $r_2 = (5.3\times 10^{-11})(2^2)/2 = (5.3\times 10^{-11})(4)/2 = 5.3\times 10^{-11} \times 2 = 1.06\times 10^{-10}\,\text{m}$. 4. Sanity check: for the $\mathrm{He}^+$ ground state ($n=1$), $r_1 = a_0/2 = 2.65\times 10^{-11}\,\text{m}$, which is SMALLER than the H ground state — a heavier nucleus pulls the electron closer. The $n = 2$ orbit being twice $a_0$ matches this trend. 5. Option A is just $a_0$ itself. Option C ignores the $/Z$ factor. Option D uses $Z = 1$. _Source: NCERT Class 12 Physics Part 2, Ch 12, §12.4 (Eq. 12.7 for orbital radii), p. 9._