A certain commercial mass spectrometer (Fig. 28-12) is used to separate uranium

Question

A certain commercial mass spectrometer (Fig. 28-12) is used to separate uranium ions of mass 3.92 x 10-25 kg and charge 3.20 x 10-19 C from related species. The ions are accelerated through a potential difference of 103 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 0.823 m. After traveling through 180° and passing through a slit of width 1.00 mm and height 0.769 cm, they are collected in a cup. (a) What is the magnitude of the (perpendicular) magnetic field in the separator? If the machine is used to separate out 1.24 mg of material per hour, calculate (b) the current (in A) of the desired ions in the machine and (c) the thermal energy (in J) produced in the cup in 1.01 h.

Explanation / Answer

Part A)

F = qvB = mv2/r

B = mv/qr

The velocity can be found such that qV = .5mv2

v = (2qV/m)

v = [(2)(3.2 X 10-19)(1.03 X 105)/(3.92 X 10-25)

v = 4.10 X 105 m/s



B = (3.92 X 10-25)(4.10 X 105)/(3.2 X 10-19)(.823)

B = .610 T



Part B)

I = Q/t

Since we have 1.24mg of material per hour, that means we can find total charge

1.24 X 10-6 kg / 3.92 X 10-25 kg/per ion shows that we have 3.16 X 1018 ions per hour

Q = (3.16 X 1018 ions )(3.20 X 10-19 C per ion)

Q = 1.012 C

I = Q/t

I = 1.012/3600

I = 2.81 X 10-4 A



Part C)

We will assume that all of the KE the ions have will be converted to heat

Therefore Q = .5mv2

Q = (.5)(1.24 X 10-6 kg/hr)(4.1 X 105 m/s)2(1.01 hr)

Q = 1.05 X 105 J

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