One way in which Uranium 235 was separated from Uranium 238 in World War 2 was t
ID: 1382757 • Letter: O
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 208 N/C for 47 centimeters in a constant magnetic field. The constant magnetic field is produced by a solenoid with 119 coils per meter wrapped around a cylinder with 7.7 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*208*0.47
= 3.1283*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*3.1283*10^-17/(3.9245*10^-25))
= 1.2626*10^4 m/s
speed of U-238, v2 = sqrt(2*KE/m1)
= sqrt(2*3.1283*10^-17/(3.9746*10^-25))
= 1.2546*10^4 m/s
Now magnetic filed produced by solenoid,
B = mue*n*I
= 4*pi*10^-7*119*7.7
= 1.15*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*1.2626*10^4/(1.15*10^-3*2*1.6*10^-19)
= 13.465 m
r2 = m2*v2/(B*q)
= 3.9746*10^-25*1.2546*10^4/(1.15*10^-3*2*1.6*10^-19)
= 13.55 m
so, d = r2 - r1
= 13.55 - 13.465
= 0.085 m or 85 cm <<<<<<<<<--------Answer
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