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alexgriva [62]
3 years ago
14

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 t

he orbit is 5.89 × 10−11 m, what is the frequency? Answer in units of rev/s
Physics
1 answer:
Tcecarenko [31]3 years ago
8 0
The attraction force provides the electron's centripetal force. 

<span>8.30^-8N = mrω² </span>

<span>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 </span>

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