Calculating Equatorial Speed of Electron Spin in Solid Sphere Model

[inferno298]

Homework Statement

When the idea of electron spin was introduced, the
electron was thought to be a tiny charged sphere (today it is
considered a point object with no extension in space). Find the
equatorial speed under the assumption that the electron is a
uniform solid sphere of radius 2.89 × 10^−6 nm, which is close
to what early theorists believed. Use the magnitude of the spin
angular momentum in your calculation.

Homework Equations

Spin angular mom = Sqrt(3/4)*hbar
L=Iw

The Attempt at a Solution

Im assuming equatorial speed is the same as angular velocity therefore I can use the moment of inertia for a solid sphere ---> I=(2/5)*m*r^2

so w=L/I
w = (Sqrt(3/4)*hbar)/((2/5)*(9.11*10^-31 kg)*(2.89*10^-15 m)
I get 3.001*10^25 m/s, which kinda seems right because that is faster than the speed of light. Which means it doesn't agree with the theory of relativity, therefore the electron must be thought of as a point charge, not a solid sphere, with spin being an intrinsic property.

I'm doing something wrong though because it isn't the right answer. Would love any help that can get me back on the right track or point out the mistake

 

[gneill]

Equatorial speed is not the same as angular velocity (although they are related). Think of the difference between the equatorial speed of the Earth due to its rotation versus the angular rotation rate of the Earth. One might be specified in m/s, the other in radians per second...

 

[inferno298]

Thank you! I got it, just had to convert my answer to linear velocity. Thanks that really helped, feel silly for forgetting that lol