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.

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Anzanzio Anzanzio · Answer 1 · 2019-12-11T17:39:08+01:00

Answer:

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