If you treat an electron as a classical rigid sphere with radius 2.00 times 10^-

Question

If you treat an electron as a classical rigid sphere with radius 2.00 times 10^-17 m and uniform density, what angular speed w is necessary to produce a spin angular momentum of magnitude squareroot3/4 h? use h = 6.63 times 10^-34 J middot 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. w = 1.22 middot 10^30 rad/s Use the equation v = 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. v = m/s

Explanation / Answer

(a)

Spin angular momentum L = I*w =  sqrt (3/4) hbar

moment of inertia l = (2/5)mr^2

I = (2/5)*9.1*10^(-31)*(2*10^(-17))^2

l = 1.45*10^(-64)

hbar = h / (2*pi) = 6.63*10^(-34) / (2*3.14) = 1.055*10^(-34)

angular speed is,

w = sqrt (3/4) hbar / l

w = sqrt(3/4)*1.05510^(-34) / (1.45*10^(-64))

w = 0.627*10^30 rad/s

(b)

v = w*r

v =  0.627*10^30 * 2*10^(-17)

v = 1.25*10^(13) m/s

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