In Bohr’s model of the hydrogen atom, what is the ratio of the magnitude of the linear momentum of the electron in the \(n=1\) orbit to that in the \(n=3\) orbit?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(3:1\)
The quantum condition for an orbit is \( mv_nr_n=nh/2\pi \), where \( mv_n \) is the linear momentum \( p_n \). The orbit radius \( r_n \) is proportional to \( n^2 \). So the momentum falls as \( 1/n \).
\( p_n=\frac{nh}{2\pi r_n} \)
\( p_n\propto\frac{n}{n^2} \)
\( \begin{aligned} \frac{p_1}{p_3} &= \frac{1/1}{1/3} \\ &= \frac{3}{1} \end{aligned} \)
From the above, \( p_1:p_3=3:1 \). The electron moves fastest in the innermost orbit.
NCERT (2026-27):
De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation – “… we have”
\( \displaystyle 2\pi r_n=\frac{nh}{mv_n}\quad\text{or}\quad \)\( \displaystyle mv_nr_n=\frac{nh}{2\pi} \)
Bohr Model of the Hydrogen Atom – “The radius of nth possible orbit thus found is”
\( \displaystyle r_n=\frac{n^2}{m}\left(\frac{h}{2\pi}\right)^2\frac{4\pi\varepsilon_0}{e^2} \)\( \qquad (12.7) \)