What is the radius of the orbit of an electron?

  • Thread starter bbbl67
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  • #1
bbbl67
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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
 

Answers and Replies

  • #2
Charles Link
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I agree with your equations, And yes, edit, I agree with your result.
 
  • #3
Borek
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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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