According to Bohr's model of the hydrogen atom,(A) the radius of the…

The correct answer is Option (2) → (B), (C) and (D) only

In the Bohr model, the electron moves in a circle under the Coulomb attraction of the nucleus, and only those orbits are allowed in which the angular momentum is quantised, \( mvr = nh/2\pi \). Solving these two conditions together gives, for hydrogen:

    \( \displaystyle r_n = \frac{\varepsilon_0 n^{2} h^{2}}{\pi m e^{2}} = 0.53\,n^{2}\,\text{angstrom} \Rightarrow r_n \propto n^{2} \)

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

    \( \displaystyle E_n = -\frac{m e^{4}}{8\varepsilon_0^{2} n^{2} h^{2}} = -\frac{13.6}{n^{2}}\,\mathrm{eV} \Rightarrow E_n \propto \frac{1}{n^{2}} \)

(A) the radius of the orbit of an electron is directly proportional to n. — Incorrect. The radius is proportional to n², not to n: the second orbit is four times as large as the first.

(B) the speed of the orbiting electron is directly proportional to 1/n — Correct. The speed is greatest in the first orbit (about 2.2 × 10⁶ m s⁻¹) and falls as 1/n in the outer orbits.

(C) the total energy of the electron is directly proportional to 1/n² — Correct. \( E_n = -13.6/n^{2} \) eV; the constant of proportionality is negative because the electron is bound, so the energy rises towards zero as n increases.

(D) the radius of the orbit of an electron is directly proportional to n² — Correct. This follows from \( r_n = 0.53\,n^{2} \) angstrom worked out above.

NCERT (2026-27):
Bohr Model of the Hydrogen Atom – “… Eq. (12.14) depicts that the orbital speed in the \( n^{\mathrm{th}} \) orbit falls by a factor of \( n \). … From Eq. (12.15), it can also be seen that the radii of the orbits increase as \( n^{2} \). The total energy of the electron in the stationary states of the hydrogen atom can be obtained by substituting the value of orbital radius in Eq. (12.4) as

    \( \displaystyle E_n = -\left(\frac{e^{2}}{8\pi\varepsilon_0}\right)\left(\frac{m}{n^{2}}\right)\left(\frac{2\pi}{h}\right)^{2}\left(\frac{e^{2}}{4\pi\varepsilon_0}\right) \)

or

    \( \displaystyle E_n = -\frac{m e^{4}}{8 n^{2}\varepsilon_0^{2} h^{2}} \qquad (12.17) \)”