If you treat an electron as a classical spherical object with radius 1.80×1017 m
ID: 1485127 • Letter: I
Question
If you treat an electron as a classical spherical object with radius 1.80×1017 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. Express your answer in radians per second to three significant figures.
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.
Express your answer in meters per second to three significant figures.
Explanation / Answer
We use the constants and:
=6.63*10^-34 / 2*pi = 1.05*10^(-34) so the mometum is
L= (1.05*10^(-34))^(1/2) *(3/4) = 7.68*10^-18
so L= I x w where I is the inercy of the ball, I= (2/5)*M*R^2 = (2/5)*(9.11×1031)*(1.80×1017)^2 = 1.179*10^-64
w=L/I = 6.51*10^(46) 1/s this is the angular speed
v=r*w= (1.80×1017) *( 6.51*10^(46)) = 1.17*10^30 m/s