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 205 N/C for 41 centimeters in a constant magnetic field. The constant magnetic field is produced by a solenoid with 164 coils per meter wrapped around a cylinder with 6.5 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

mass of uranium-235, m1 = 235*1.67*10^-27

= 3.9245*10^-25 kg

mass of uranium-238, m2 = 238*1.67*10^-27

= 3.9746*10^-25 kg

Workdone on the isotopes, W = q*E*d

= 2*1.6*10^-19*205*0.41

= 2.6896*10^-17 J

By Applying work-enrgy theorem,

W = 0.5*m*v^2

==> v = sqrt(2*Ke/m)

so,

speed of U-235, v1 = sqrt(2*KE/m1)

= sqrt(2*2.6896*10^-17/(3.9245*10^-25))

= 1.17*10^4 m/s

speed of U-238, v2 = sqrt(2*KE/m1)

= sqrt(2*2.6896*10^-17/(3.9746*10^-25))

= 1.163*10^4 m/s

Now magnetic filed produced by solenoid,

B = mue*n*I

= 4*pi*10^-7*164*6.5

= 1.34*10^-3 T

now Apply, m*v^2/r = q*v*B

==> r = m*v/(B*q)

r1 = m1*v1/(B*q)

= 3.9245*10^-25*1.17*10^4/(1.34*10^-3*2*1.6*10^-19)

= 10.7 m

r2 = m2*v2/(B*q)

= 3.9746*10^-25*1.163*10^4/(1.34*10^-3*2*1.6*10^-19)

= 10.78 m

so, d = r2 - r1

= 10.78 - 10.7

= 0.08 m or 8 cm <<<<<<<<<--------Answer

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