For a dynamically stable orbit in a hydrogen atom, according to Rutherford’s atomic model the relation between the orbit radius (\( r \)) and the electron velocity (\( v \)) is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( r = \displaystyle\frac{e^2}{4\pi\varepsilon_0 m v^2} \)
For a dynamically stable orbit, the electrostatic attraction between the proton and the electron provides the centripetal force needed for circular motion.
\( \displaystyle \frac{m v^{2}}{r} = \frac{1}{4\pi\varepsilon_0}\,\frac{e \cdot e}{r^{2}} \)
Multiplying both sides by \( r^{2} \) and dividing by \( m v^{2} \):
\( \displaystyle r = \frac{e^{2}}{4\pi\varepsilon_0\,m v^{2}} \)
NCERT (2026-27):
Electron Orbits – “… The electrostatic force of attraction, Fe between the revolving electrons and the nucleus provides the requisite centripetal force (Fc) to keep them in their orbits. Thus, for a dynamically stable orbit in a hydrogen atom
\( \displaystyle F_e = F_c \)
\( \displaystyle \frac{m v^2}{r} = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} \qquad (12.2) \)
Thus the relation between the orbit radius and the electron velocity is
\( \displaystyle r = \frac{e^2}{4\pi\varepsilon_0 m v^2} \qquad (12.3) \)”