A) If you treat an electron as a classical spherical object with radius 1.90×10

Question

A) If you treat an electron as a classical spherical object with radius 1.90×1017 m , what angular speed is necessary to produce a spin angular momentum of magnitude 3/4? Use h = 6.63×1034 Js for Planck's constant, recalling that =h/2, and 9.11×1031 kg for the mass of an electron.

answer in rad/s.

B) Use the equation v=r relating velocity to radius and angular velocity together with the result of Part A to calculate the speed v of a point at the electron's equator. answer in m/s.

please explain. I appreciate it very much!

Explanation / Answer

Moment of inertia of the solid sphere is I = (2/5)*m*r^2

Spin angular momentum L = I*w

But given that L = sqrt(3/4)hbar

sqrt(3/4)hbar = (2/5)*m*r^2*w

angular speed is w = sqrt(3/4)hbar/((2/5)*m*r^2)

hbar = h/(2*pi) = 6.63*10^-34/(2*3.142) = 1.055e-34

I = (2/5)*9.11*10^-31*(1.9*10^-17)^2 = 1.31e-64

then w = sqrt(3/4)*1.055e-34/(1.31e-64) = 6.97*10^29 rad/s

B) v = r*w = 1.9*10^-17*6.97*10^28 = 1.324e+12 m/s

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