One way in which Uranium 235 was separated from Uranium 238 in World War 2 was t

Question

One way in which Uranium 235 was separated from Uranium 238 in World War 2 was to use a mass spectrograph to separate the 2 isotopes. Suppose the uranium atoms are ionized so that they have a net charge of +2 elemental charge units. These ions are then accelerated through a constant electric field of 202 N/C for 52 centimeters in a constant magnetic field. The constant magnetic field is produced by a solenoid with 147 coils per meter wrapped around a cylinder with 9.1 Amps of current running through it. The two ions have different paths through the magnetic field into 2 different containers as in the diagram below. How far apart, d in centimeters, do the centers of the containers have to be to maximize collecting both isotopes?

Explanation / Answer

First, by conservation of energy, PE = KE

qEd = .5mv2

mass of Uranium 235 = 235(1.6605 X 10-27) = 3.902 X 10-25 kg

mass of Uranium 238 = 238(1.6605 X 10-27) = 3.952 X 10-25 kg

Find the velocity of each...

(2)(1.6 X 10-19)(202)(.52) = .5(3.902 X 10-25)(v2)

U 235 v = 13126.7374 m/s

(2)(1.6 X 10-19)(202)(.52) = .5(3.952 X 10-25)(v2)

U 238 v = 13402.4408 m/s

Now we need the B field

B = uNI/L

B = (4pi X 10-7)(147)(9.1)/1

B = .001681 T

r = mv/qB

For U 235

r = (3.902 X 10-25)(13126.7374)/(2)(1.6 X 10-19)(.001681)

r = 9.5220 m

For U 238

r = (3.952 X 10-25)(13402.4408)/(2)(1.6 X 10-19)(.001681)

r = 9.8465 m

Subtract the two radii

9.8465 - 9.5220 = .3245 m

d = .3245 m which is 32.45 cm

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