In a muonic hydrogen atom, a muon (which has a mass 200 times that...

For a particle of mass $m$ and charge $e$ revolving around a proton with speed $v$ in an orbit of radius $r$, the centripetal force is provided by the electrostatic force of attraction.

$\dfrac{mv^2}{r} = \dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{r^2}$

Rearranging for the radius $r$:

$r = \dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{mv^2}$

For a fixed orbital speed $v$, the radius $r$ is inversely proportional to the mass $m$ of the revolving particle ($r \propto \dfrac{1}{m}$).

Therefore, the ratio of the radii is:

$\dfrac{r_{\mu}}{r_e} = \dfrac{m_e}{m_{\mu}} = \dfrac{m_e}{200 m_e} = \dfrac{1}{200}$

The ratio is $1 : 200$.

Answer: $1 : 200$