The radius of \( n^{\mathrm{th}} \) orbit of H-atom is ‘\( r \)’, then \( r \) and \( n \) are related as
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( r \propto n^{2} \)
In the Bohr model, the quantization of angular momentum, \( mvr = nh/2\pi \), together with the Coulomb attraction providing the centripetal force, gives the radius of orbit n of hydrogen:
\( \displaystyle r_n = \frac{\varepsilon_0 n^{2}h^{2}}{\pi me^{2}} = (0.529\times10^{-10}\,\mathrm{m})\,n^{2} \)
So \( r \propto n^{2} \): the radius grows as the square of the orbit number.
Why the other options are incorrect
Option (1) \( r \propto n \) — Incorrect. The radius goes as \( n^{2} \), not n; the quantity proportional to n is the angular momentum.
Option (3) \( r \propto \displaystyle\frac{1}{n} \) — Incorrect. It is the speed of the electron that varies as \( 1/n \), not the radius.
Option (4) \( r \propto \displaystyle\frac{1}{n^{2}} \) — Incorrect. It is the magnitude of the energy that varies as \( 1/n^{2} \), not the radius.
NCERT (2026-27):
Bohr Model of the Hydrogen Atom – “… For a hydrogen atom, Eq. (12.4) gives the expression to determine the energies of different energy states. But then this equation requires the radius \( r \) of the electron orbit. To calculate \( r \), Bohr’s second postulate about the angular momentum of the electron–the quantisation condition – is used. The radius of \( n^{\mathrm{th}} \) possible orbit thus found is
\( \displaystyle r_n = \left(\frac{n^{2}}{m}\right)\left(\frac{h}{2\pi}\right)^{2}\frac{4\pi\varepsilon_0}{e^{2}} \qquad (12.7) \)”