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Answer: \(r_n = \dfrac{\varepsilon_0 h^2 n^2}{\pi m Z e^2}\) (for hydrogen \(Z = 1\)).

Consider an electron of mass \(m\), charge \(e\) moving with speed \(v_n\) in the \(n\)th circular orbit of radius \(r_n\) around a nucleus of charge \(Ze\).

(1) The electrostatic attraction provides the centripetal force:

\[\frac{m v_n^2}{r_n} = \frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r_n^2} \quad\Rightarrow\quad m v_n^2 r_n = \frac{Ze^2}{4\pi\varepsilon_0} \qquad (1)\]

(2) Bohr’s quantum condition: angular momentum is an integral multiple of \(h/2\pi\):

\[m v_n r_n = \frac{nh}{2\pi} \quad\Rightarrow\quad v_n = \frac{nh}{2\pi m r_n} \qquad (2)\]

Substituting (2) in (1):

\[m\left(\frac{nh}{2\pi m r_n}\right)^2 r_n = \frac{Ze^2}{4\pi\varepsilon_0} \Rightarrow \frac{n^2h^2}{4\pi^2 m r_n} = \frac{Ze^2}{4\pi\varepsilon_0}\]\[\boxed{r_n = \frac{\varepsilon_0 h^2 n^2}{\pi m Z e^2}}\]

So \(r_n \propto n^2\). For hydrogen in the ground state (\(n = 1\), \(Z = 1\)) this gives the Bohr radius \(r_1 \approx 0.53\) Å.

Doubtly AI · 27 Oct 2026 AI-verified · Oct 2026 Report
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