In a hydrogen atom, the electron revolves round the nucleus in a circular orbit of radius \(r\) with speed \(v\). If the electron moves in an orbit of radius \(9r\), its speed will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(v/3\)
The electrostatic attraction of the nucleus provides the centripetal force for the electron. Equating the two forces gives a relation between the orbit radius and the speed.
\( \displaystyle r=\frac{e^{2}}{4\pi\varepsilon_0mv^{2}} \)
So \( v^{2}\propto 1/r \), or \( v\propto 1/\sqrt{r} \).
\( \begin{aligned} \frac{v^{\prime}}{v} &= \sqrt{\frac{r}{9r}} \\ &= \frac{1}{3} \end{aligned} \)
From the above, \( v^{\prime}=v/3 \). Hence, the speed in the larger orbit is one-third of the speed in the first orbit.
NCERT (2026-27):
Electron orbits – “Thus the relation between the orbit radius and the electron velocity is”
\( \displaystyle r=\frac{e^{2}}{4\pi\varepsilon_0mv^{2}} \)\( \qquad (12.3) \)