Bohr Model

Bohr postulated that only orbits in which the electron's angular momentum is an integral multiple of $\frac{h}{2\pi }$ are allowed, and from this quantisation NCERT derives the orbit radius $r_n=\frac{n^2{h}^2{\epsilon }_0}{\pi m{e}^2}=n^2{a}_0$, where $a_0=0.053$ nm is the Bohr radius. The radius therefore scales as the square of the principal quantum number. For the second orbit $n=2$, so $r_2=2^2\times 0.053=4\times 0.053=0.212$ nm. The choice 0.053 nm is wrong because that is the ground-state radius for $n=1$ and ignores the $n^2$ scaling. The choice 0.106 nm is wrong because it multiplies the Bohr radius by $n$ rather than $n^2$, a common linear-scaling error. The choice 0.477 nm is wrong because it corresponds to $n=3$ ($9a_0$) and not the second orbit. As stated in NCERT Class 12, Chapter 12 (Atoms), the discrete radii grow rapidly with $n$, so the electron is far weaker bound in outer orbits. A magnitude check confirms that quadrupling the ground radius gives roughly a fifth of a nanometre, consistent with atomic dimensions.