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In a hydrogen atom, the electron in orbit around the proton feels an attractive force of about 8.54 × 10−8 N. If the radius of the orbit is 5.82 × 10−11 m, what is the frequency? Answer in units of rev/s

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\[F_c=\frac{m_ev^2}{r}=m_e\omega^2\\\omega=\sqrt{\frac{F_c}{m_e}}\\2\pi f=\sqrt{\frac{F_c}{m_e}}\\f=\text{...?} \]
Consider what kind of acceleration this particle is undergoing. a=v^2/r for centripetal acceleration. Now to know how many revolutions, consider, what is the distance of one revolution? It's the same as the circumference of a circle with a radius the same as the electron. Boom, now all you need to do is put it all together with formulas you know. Remember, you can use "revolutions" as a unit like "meters" or "newtons".
^ fail ^

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thank u all

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