The speed of an electron in an orbit in a hydrogen atom is proportion…

The correct answer is Option (3) → n⁻¹

In Bohr’s model, angular momentum is quantised, \( mvr = nh/2\pi \), and the orbit radius grows as \( n^{2} \).

    \( \displaystyle r_n \propto n^{2} \)

    \( \displaystyle v_n = \frac{nh}{2\pi m r_n} \propto \frac{n}{n^{2}} \)

    \( \displaystyle v_n \propto \frac{1}{n} = n^{-1} \)

For hydrogen, \( v_n = \frac{2.19\times10^{6}}{n}\,\mathrm{m\,s^{-1}} \).

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), but it is still in the CUET 2026 syllabus, so do not skip it.

NCERT (Pre-rationalised edition, 2022-23):
Bohr Model of the Hydrogen Atom – “… Combining it with Eq. (12.13), we get the following expressions for v_n and r_n,

    \( \displaystyle v_n = \frac{1}{n}\,\frac{e^2}{4\pi\varepsilon_0}\,\frac{1}{(h/2\pi)} \qquad (12.14) \)

and

    \( \displaystyle r_n = \left(\frac{n^2}{m}\right)\left(\frac{h}{2\pi}\right)^2 \frac{4\pi\varepsilon_0}{e^2} \qquad (12.15) \)

Eq. (12.14) depicts that the orbital speed in the nth orbit falls by a factor of n.”