If you treat an electron as a classical spherical object with radius 2.00 Times
ID: 1479539 • Letter: I
Question
If you treat an electron as a classical spherical object with radius 2.00 Times 10^-17 m , what angular speed omega is necessary to produce a spin angular momentum of magnitude square root 3/4 ? Use h = 6.63 Times 10^-34 J s for Planck's constant, recalling that h = h/2 pi, and 9.11 Times 10^-31 kg for the mass of an electron. Express your answer in radians per second to three significant figures. Use the equation v = r omega 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
A)
I of sphere = (2/5)*m*r^2
= (2/5)*(9.11*10^-31)*(2*10^-17)^2
= 1.46*10^-64 Kg.m^2
Angular momentum,
L =sqrt(3/4)* h/2*Pi
= sqrt (3/4) * (6.626*10^-34)/ (2*pi)
= 9.13*10^-35 Kgm^2/s
L = I*w
9.13*10^-35 = 1.46*10^-64 * w
w = 6.3*10^29 rad/s
Answer: 6.3*10^29 rad/s
B)
use:
V = r* w
=(2*10^-17)*(6.3*10^29)
= 1.25*10^13 m/s
Answer: 1.25*10^13 m/s