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Rotation period of electron orbiting a proton

  • Thread starter SorenaJ
  • Start date
5
0
1. Homework Statement
An electron is rotating around a proton (at rest) in a perfect circular orbit. If the radius of the orbit is r=10^-10 m, how long is the rotation period [hint: the radial acceleration is entirely due to the electric force]

k=9*10^9
q=1.6*10^-19

2. Homework Equations
1. F=(k*q^2)/r^2
2. arad=(angular velocity)^2*r


3. The Attempt at a Solution
I found the force by equaiton 1., and I got 2.3*10^-8 N ((9*10^9)(1.6*10^-19)^2))/((10^-10)^2)
I tried equation 2. to get (angular velocity)=sqrt(F/r) and got 15.2 rad/s
This means that it is 2.4 rev/s (by dividing by 2pi) and 0.41 seconds per orbital period.
That is like the world's slowest electron. Where did I go wrong?
 

lightgrav

Homework Helper
1,248
30
Is Force equal to acceleration? ... or is there something missing?
always use units, to avoid slip-ups like this.
 
5
0
Maybe not. It just says that the radial acceleration is entirely due to the electric force. To find the radial acceleration do I just need to say a=F/m?
 

lightgrav

Homework Helper
1,248
30
well, the radial acceleration is caused entirely by the electric force, but it is mitigated via (spread thru-out) the electron mass
= .911×10-30 kg .
 

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