Consider the atomic model of hydrogen atom given by Niels Bohr. n is…

The correct answer is Option (4) → A - III, B - IV, C - II, D - I

Bohr’s results for the hydrogen atom follow from the quantisation of angular momentum together with the Coulomb force providing the centripetal force.

(A) Angular momentum of the electron is proportional to — this matches (III) n. By Bohr’s quantisation condition the angular momentum is a whole multiple of \( h/2\pi \).

    \( \displaystyle L = \frac{nh}{2\pi} \)

(B) Velocity of an electron in an orbit is proportional to — this matches (IV) 1/n. The electron moves more slowly in outer orbits.

    \( \displaystyle v_n = \frac{e^{2}}{2\varepsilon_0 h}\cdot\frac{1}{n} \)

(C) Radius of the electron orbit is proportional to — this matches (II) n². The orbit radius grows as the square of n.

    \( \displaystyle r_n = \frac{\varepsilon_0 h^{2}}{\pi m e^{2}}\,n^{2} \)

(D) Energy of the electron in an orbit is proportional to — this matches (I) 1/n². The energy is negative and its magnitude falls as \( 1/n^{2} \).

    \( \displaystyle E_n = -\frac{13.6}{n^{2}}\,\mathrm{eV} \)

NCERT (2026-27):
Summary – “… (b) The stationary orbits are those for which the angular momentum is some integral multiple of h/2π. (Bohr’s quantisation condition.) That is \( L = nh/2\pi \), where n is an integer called the principal quantum number.”

Summary – “… 7. As a result of the quantisation condition of angular momentum, the electron orbits the nucleus at only specific radii. For a hydrogen atom it is given by

    \( \displaystyle r_n = \left(\frac{n^2}{m}\right)\left(\frac{h}{2\pi}\right)^2 \frac{4\pi\varepsilon_0}{e^2} \)

The total energy is also quantised:

    \( \displaystyle E_n = -\frac{m e^4}{8 n^2 \varepsilon_0^2 h^2} = -13.6\ \mathrm{eV}/n^2 \)”