What is the radius of the orbit of an electron?

AI Thread Summary
The discussion centers on calculating the radius of an electron's orbit around a zinc nucleus using classical physics principles. The approach involves balancing centripetal force and Coulomb's Law to derive the radius formula. The specific values used include the charge of the electron, the total charge of the zinc nucleus, the mass of the electron, and its velocity. The calculated radius is approximately 93.8 picometers. Quantum mechanics is noted as not applicable for discussing orbits or electron velocity in this context.
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Homework Statement


What is the radius of the orbit of an electron traveling at 9.0 x 10^6 m/s around a zinc nucleus which contains 30 protons?

Homework Equations


I don't know if this problem can be solved quantum mechanically, all I can think of doing is solving it classically, using Coulomb's Law and Centripetal force.

(1) Centripetal Force:
F = (m v^2)/r |
r | radius
F | centripetal force
v | rotation speed
m | mass
https://is.gd/CL6WBe

(2) Coulomb's Law:
F = (Q_1 Q_2)/(4 pi e_0 r^2) | U = (Q_1 Q_2)/(4 pi e_0 r) |
F | force
Q_1 | charge 1
Q_2 | charge 2
r | distance
U | potential energy
e_0 | electric constant (˜ 8.854×10^-12 F/m)
(in vacuum)
https://is.gd/P5JR1H

The Attempt at a Solution


Then you use the balance of forces to find the radius.
F(1) = F(2)
(m v^2)/r = (Q_1 Q_2)/(4 pi e_0 r^2)
r = (Q_1 Q_2)/(4 pi e_0 m v^2) = ?
Q_1 = 1 e = 1.6021766×10^-19 C
Q_2 = 30 e = 4.8065299×10^-18 C
m = 1 m_e = 9.109384×10^-31 kg
v = 9.0E+6 m/s
r = ( 1.6021766×10^-19 C * 4.8065299×10^-18 C) / ( 4 pi e_0 (9.109384×10^-31 kg) (9.0E+6 m/s)^2)
= 9.380143×10^-11 meters
= 93.80143 picometers
https://is.gd/ttIdyJ
 
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I agree with your equations, And yes, edit, I agree with your result.
 
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This calls for LaTeX.

In QM you can't talk about an orbit nor the electron velocity, so QM would be of no help here.

Other than that I trust Charles :wink:
 
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