Revolutions per second question

[guru]

An electron of mass (m=9.1*10^(-31)kg) orbits a proton at a distance of 5.3*10^(-11)m. The proton pulls on the electron with an electric force of 9.2*10^(-8)N.
How many revolutions per second does the electron make?

This is what I did:

m=9.1*10^(-31)kg
v=5.3*10^(-11)m
T=9.2*10^(-8)N

V=sqrt(rT/m)= 23147876.27 m/s

angular velocity (w) = (v/r) = 4.38^(16) rad/s
w*(1 rev/(2(pi)rad)) = 9.17*10^9 rev/s

I was told the answer was wrong.
Please help

 

[gnome]

First, I think you have a decimal-place error in your tangential velocity calculation.

Anyway...

$$v = \sqrt{ \frac{T*r}{m}}$$
(I'm following your lead in calling the force T even though it's not a tension force.)

So we can use F for the frequency in revolutions per second
and C is the circumference of the orbit.
The distance traveled divided by the speed gives you the time it takes to cover that distance, so the inverse of that (the speed divided by the distance) will give you the number of revolutions per second.
$$F = \frac{v}{C}$$

$$F = \frac{\sqrt{\frac{T*r}{m}}}{2\pi r} = \frac{1}{2\pi} \times \sqrt{ \frac{T}{mr}}$$

 

[guru]

Got it!
Thanks a lot