In a hydrogen atom, the electron in orbit around the proton feels an attractive force of about 7.45 * 10-8 N. If the radius of the orbit is 5.89 * 10-11 m, what is the frequency? Answer in units of rev/s

The attraction force provides the electron's centripetal force.

8.30^-8N = mrω²

ω² = 8.30^-8 / (9.11^-31 kg x 4.70^-11m) ... ω² = 1.94^33 (rad/s) ² ... ω = 4.40^16 rad/s

f = ω/2π = 4.40^16 / 2π ... ►f = 7.0^15 Hz