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 x 1o25 kg and charge 3.20x 1019 C from related species. The ions are accelerated through a potential difference of 114 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 0.776 m. After traveling through 180° and passing through a slit of width 0.784 mm and height 0.844 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 0.962 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 1.30 h Detector (a) Number (b) Number (c) Number Units Units Units

Explanation / Answer

A particle with charge z, mass m, and speed v moving perpendicular to a magnetic field B orbiting in a circular path of radius r given by:

r = m*v/(B*z)

B = m*v/(r*z)

A charged particle accelerated through a voltage V has a kinetic energy equal to:

(1/2)*m*v^2 = z*V

so the speed of the accelerated particle is:

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

Plug this into the expression above for the magnetic field to get:
B = sqrt(2*m*V/(z*r^2))
In this case,
V = 114*10^3 volts
m = (235 gm/mol)/(6.022*10^23 atoms/mol) = 3.92*10^-25 kg
z = 3.2*10^-19 C
r = 0.776 m
Plugging all this into the equation for B, we find that:
B = B = sqrt(2*3.92*10^-25*114*10^3/(3.2*10^-19*.776^2))= 0.68 T

b)In this we need to calculate the number of ions reaching the detector per second:

0.962mg/hr = 0.962*10^{-3} gm/hr
(0.962 *10^{-3}gm/hr)/(235 gm/mol) = 4.1*10 ^-6 mol/hr

(4.1*10^-6 mol/hr)*6.022*10^23 ions/mol = 24.69*10^17 ions/hr

(24.69*10^17 ions/hr) * (1/60^2 hr/sec) * (2 elementary charges/ion) * (1.602*10^-19 C/elem charge) = 2.19*10^-4 A
c)In this we'll ignore any resistive heating of the Faraday cup, and assume that all the heating is caused by the deposition of the kinetic energy of the incoming ions.

The kinetic energy of each ion is given by:

z*V = (2*1.602*10^-19 C/ion)*114*10^3 V = 3.648*10^-14 J/ion
We already calculated that there are
(24.69*10^17 ions/hr) = 4.115*10^16 ions/sec reaching the collector, so the total energy being dissipated (as heat) in the collector per unit time is:

4.115*10^16 ions/sec * 3.648*10^-14 J/ion = 1.5*10^3 W

So in 1.3hr=1.3*60sec it produce 1.5*10^3 *1.3*60=117*10^3 J

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