Bohr around a wire

First, we get the electric field around the wire, which is easy using Gauss' flux law

$\oint \vec{E}\cdot\vec{dA} = \frac{\lambda l}{\varepsilon_0} \rightarrow E = \frac{\lambda}{\varepsilon_0 2\pi r_\text{orbit}}$

even when $l$ goes to infinity.

The assumption we make is that the electron is kept in orbit as are satellites, through a balance of centripetal acceleration and electric attraction. Therefore, we have

$m_ev^2/r_\text{orbit} = \frac{e\lambda}{\varepsilon_0 2\pi r_\text{orbit}}$

which leads to $\displaystyle v = \sqrt{\frac{e\lambda}{\varepsilon_0 2\pi m_e}}$ .

The key to finishing this off is the assumption of Bohr, which is that the angular momentum of a particle can only increment by integer multiples of $\hbar$ , i.e. $L = m_evr_\text{orbit} = n\hbar$ . We've already shown that $v$ is independent of the radius of orbit, so $r_\text{orbit}$ should be minimized by taking the smallest angular momentum, $n=1$ .

We have $m_e v r=\hbar$ , thus

$\begin{aligned} m_e r v &= \hbar \\ m_e^2r^2v^2 &=\hbar^2\end{aligned}$

From here, it is important to remember that we are only dealing in magnitudes, due to the definition of $\lambda$ in the problem. Reading these expressions literally will result in incorrect units.

$\begin{aligned} r^2 &= \bigl| \hbar^2\frac{\varepsilon_0 2\pi}{e\lambda m_e} \bigr| \\ r^2 &= \bigl|\hbar^2 \frac{\varepsilon_0 2\pi}{e m_e} \frac{200\pi m_e e}{\varepsilon_0} \bigr| \\ r^2 &=\bigl|\hbar^2 2\pi\times 200\pi\bigr| \\ r^2 &= \bigl| 100h^2\bigr| \end{aligned}$

which shows that $r=10\lvert h\rvert$ m. $\approx 6.636\text{E}-33$ m.