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

Question

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

anonymous · Answer

$$F_c=\frac{m_ev^2}{r}=m_e\omega^2\\omega=\sqrt{\frac{F_c}{m_e}}\2\pi f=\sqrt{\frac{F_c}{m_e}}\f=\text{...?}  $$

kainui · Answer

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".

anonymous · Answer

^ fail ^

anonymous · Answer

thank u all