A certain commercial mass spectrometer is used to separate uranium ions of mass

Question

A certain commercial mass spectrometer is used to separate uranium ions of mass 3.92 times 10^-25 kg and charge 3.20 times 10^-19 C from related species. The ions are accelerated through a potential difference of 132 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 1 38 m. After traveling through 180 degree and passing through a slit of width 0.858 mm and height 1.09 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 0.838 mg of material per hour, calculate (b) the current of the desired ions in the machine and (c) the thermal energy produced in the cup in 0.953 h.

Explanation / Answer

Speed of the accelerated particle is:

v = sqrt(2*z*V/m)

So,

B = sqrt(2*m*V/(z*r^2))
Given,

V = 132000 volts

m = 3.92*10^-25 kg

z = 2*1.602*10^-19

C = 3.20*10^-19 C

r = 1.38 m

B = sqrt(2*m*V/(z*r^2))
=> B = sqrt(2*3.92*10^-25*132000/(2*1.602*10^-19*1.38^2))

B = 0.412 T.

For part (b),

0.838mg/hr = 0.000838 gm/hr

(0.000838 mg/hr)/(235 mg/mol) = 3.6*10 ^-6 mol/hr

(3.6*10 ^-6 mol/hr)*6.022*10^23 ions/mol = 2.17*10^18 ions/hr

(2.17*10^18 ions/hr) * (1/60^2 hr/sec) * (2 elementary charges/ion) * (1.602*10^-19 C/elem charge) = 1.93*10^-3 A

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