If you treat an electron as a classical rigid sphere with radius 1.70*10-17 m and uniform density, what angular speed ω is necessary to produce a spin angular momentum of magnitude 3/4---√ℏ? Use h = 6.63*10-34 J⋅s for Planck's constant, recalling that ℏ=h/2π, and 9.11*10-31 kg for the mass of an electron. Express your answer in radians per second to three significant figures.

i) The angular speed is ω = 8.68*10²⁹ rad/s

Explanation:

Moments of inertia of solid sphere with radius r:

I = (2/5) mr²

The expression for angular momentum is:

L = Iω = (2/5) mr²ω = sqrt (3/4) ћ

Rearranging the above equation to make ω the subject of the formula, we get:

ω = sqrt (3/4) ћ / (2/5) mr²

where

ћ = h/2π = (6.63*10-34 J⋅s) / 2π = 1.055*10⁻³⁴

Thus,

ω = [ (1.055*10⁻³⁴) √ (3/4) ]/2/5 (9.11*10⁻³¹) (1.70*10⁻¹⁷) ²

ω = 8.68*10²⁹ rad/s