Arrange the speeds of the electron revolving in the orbits (characterized by their principal quantum number (n) values) of a hydrogen atom, in ascending order.
(A) n = 2
(B) n = 4
(C) n = 1
(D) n = 3
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (B), (D), (A), (C)
In the Bohr model the speed of the electron in the n-th orbit of hydrogen is inversely proportional to n:
\( \displaystyle v_n = \frac{e^{2}}{2\varepsilon_0 h}\,\frac{1}{n} \approx \frac{2.18\times10^{6}}{n}\,\mathrm{m\,s^{-1}} \)
So the speeds are in the ratio
\( v_4 : v_3 : v_2 : v_1 = \tfrac{1}{4} : \tfrac{1}{3} : \tfrac{1}{2} : 1 \)
The electron is slowest in the outermost orbit, so the ascending order is n = 4, 3, 2, 1, i.e. (B), (D), (A), (C).
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 vn and rn,
\( \displaystyle v_n = \frac{1}{n}\,\frac{e^2}{4\pi\varepsilon_0}\,\frac{1}{(h/2\pi)} \qquad (12.14) \)”
Bohr Model of the Hydrogen Atom – “… Eq. (12.14) depicts that the orbital speed in the nth orbit falls by a factor of n.”