If you treat an electron as a classical spherical object with radius 1.30 times

Question

If you treat an electron as a classical spherical object with radius 1.30 times 1-17m , what angular speed omega is necessary to produce a spin angular momentum of magnitude Use h = 6.63 times 10-34J s for Planck's constant, recalling that = 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

Part A:

angular momentum. L = (3/4)*(h/2) = (3/4)*(6.63*10-34)/(2) =9.138*10-35 kg-m2/s

electron is assumed as spherical.

therefore, moment of inertia of the electron, I= 2mR2/5 = 2*9.11*10-31*(1.3*10-17)2/5 = 6.158*10-65 kg-m2

angualar momentum, L = I

therefore, angular velocity, = L/I = (9.138*10-35)/(6.158*10-65) = 1.484*1030 rad/s

Part B:

v=r

v= (1.3*10-17)*(1.484*1030)= 1.929*1013 m/s

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