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(A) The frequency of revolution of the electron is given by $f = \frac{v}{2 \pi r}$.Substituting the given values: $f = \frac{2.2 	imes 10^{6}}{2 \pi

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$1.12$

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(A) The frequency of revolution of the electron is given by $f = \frac{v}{2 \pi r}$.Substituting the given values: $f = \frac{2.2 	imes 10^{6}}{2 \pi (5 	imes 10^{-11})} \approx 7.0 	imes 10^{15} \, Hz$.The current $i$ associated with the motion of the electron is $i = qf$, where $q$ is the charge of the electron $(1.6 	imes 10^{-19} \, C)$.$i = (1.6 	imes 10^{-19} \, C) 	imes (7.0 	imes 10^{15} \, Hz) = 11.2 	imes 10^{-4} \, A$.Converting to milliamperes: $i = 1.12 	imes 10^{-3} \, A = 1.12 \, mA$.

In the Bohr model of the hydrogen atom, the electron is pictured to rotate in a circular orbit of radius $5 \times 10^{-11} \, m$, at a speed of $2.2 \times 10^{6} \, m/s$. What is the current associated with the electron's motion (in $ \, mA$)?

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