In the \(n=2\) orbit of a hydrogen atom, the radius of the electron orbit is \(2.12\times10^{-10}\ \mathrm{m}\). The speed of the electron in this orbit is about (Take \(1/4\pi\varepsilon_0=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\), \(e=1.6\times10^{-19}\ \mathrm{C}\) and \(m=9.1\times10^{-31}\ \mathrm{kg}\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(1.1\times10^{6}\ \mathrm{m\,s^{-1}}\)
For a stable orbit, the electrostatic attraction of the proton gives the centripetal force on the electron. Setting the two forces equal gives \( r=e^{2}/4\pi\varepsilon_0mv^{2} \). So the speed can be found from the radius.
\( \begin{aligned} v^{2} &= \frac{e^{2}}{4\pi\varepsilon_0mr} \\ &= \frac{9\times10^{9}\times(1.6\times10^{-19})^{2}}{9.1\times10^{-31}\times2.12\times10^{-10}} \\ &= 1.19\times10^{12}\ \mathrm{m^{2}\,s^{-2}} \end{aligned} \)
\( \begin{aligned} v &= \sqrt{1.19\times10^{12}} \\ &= 1.1\times10^{6}\ \mathrm{m\,s^{-1}} \end{aligned} \)
This is half the speed of the electron in the ground state, \( 2.2\times10^{6}\ \mathrm{m\,s^{-1}} \).
NCERT (2026-27):
Electron orbits – “… Thus the relation between the orbit radius and the electron velocity is”
\( \displaystyle r=\frac{e^{2}}{4\pi\varepsilon_0mv^{2}} \)\( \qquad (12.3) \)