Question

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

Answer 1

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