Homework 10 Classical Electron Spin 4 of 13 > ParI A If you treat an electron as a classical rigid sphere with radius 2.00x10-17 Im and uniform density, what angular speed w is necessary to produce a spin angular momentum of magnitude v/3/4 h? Use h = 6.63x10-34 J s for Planck\'s constant, recalling that . h/2T, and 9 11x10-31 kg for the mass of an electron. Express your answer in radians per second to three significant figures View Available Hint(s) rad/s Submit PartB Use the equation u rw 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 Express your answer in meters per second to three significant figures.
Solution
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) = 1.055*10^-34
I = (2/5)*9.11*10^-31*(2*10^-17)^2 = 1.45*10^-64
then w = sqrt(3/4)*1.055*10^-34/(1.45*10^-64) = 6.30*10^29 rad/s
b.)
v = r*w = 2.0*10^-17*6.3*10^29 = 1.26*10^13 m/s
.
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Homework 10 Classical Electron Spin 4 of 13 - ParI A If you treat an e.docx
1. Homework 10 Classical Electron Spin 4 of 13 > ParI A If you treat an electron as a classical
rigid sphere with radius 2.00x10-17 Im and uniform density, what angular speed w is necessary
to produce a spin angular momentum of magnitude v/3/4 h? Use h = 6.63x10-34 J s for Planck's
constant, recalling that . h/2T, and 9 11x10-31 kg for the mass of an electron. Express your
answer in radians per second to three significant figures View Available Hint(s) rad/s Submit
PartB Use the equation u rw 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 Express your answer
in meters per second to three significant figures.
Solution
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) = 1.055*10^-34
I = (2/5)*9.11*10^-31*(2*10^-17)^2 = 1.45*10^-64
then w = sqrt(3/4)*1.055*10^-34/(1.45*10^-64) = 6.30*10^29 rad/s
b.)
v = r*w = 2.0*10^-17*6.3*10^29 = 1.26*10^13 m/s
