ions are accelerated through a potential difference of 100 kV and then pass into

Question

ions are accelerated through a potential difference of 100 kV and then pass into a uniform magnetic field

where they are deflected into a circular path of radius 0.90 m. After travelling through 180 degrees, the beam is

collimated and collected in a Faraday cup.

(a) What is the magnitude of the (perpendicular) magnetic field in the separator?

If 100mg of material is separated per hour, calculate (b) the current of the desired ions in the machine,

and (c) the rate at which heat is produced in the cup i.e. the rate of loss of kinetic energy of the ions collected

in the Faraday cup

Explanation / Answer

A commercial mass spectrometer is used to separate Uranium235 (++) ions from ions of other uranium isotopes.The ions are accelerated through a potential difference of 100 kV and then pass into a uniform magnetic field where they are deflected into a circular path of radius 1.00m. After travelling through 180 degrees, the beam is collimated and collected in a Faraday cup.

(a) What is the magnitude of the (perpendicular) magnetic field in the separator? If 100mg of material is separated per hour, calculate

(b) the current of the desired ions in the machine,and

(c) the rate at which heat is produced in the cup i.e. the rate of loss of kinetic energy of the ions collected in the Faraday cup

A particle with charge z, mass m, and speed v moving perpendicular to a magnetic field B will orbit 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 = 10^5 volts

m = (235 gm/mol)/(6.022*10^23 atoms/mol) = 3.902*10^25 kg

z = 2*1.602*10^-19 C = 3.204*10^-19 C

r = 1 m

Plugging all this into the equation for B, we find that:

B = 0.494 T. This is the answer to part (a).

For part (b), we need to calculate the number of ions reaching the detector per second:

100mg/hr = 0.1 gm/hr

(0.1gm/hr)/(235 gm/mol) = 4.255*10 ^-4 mol/hr

(4.255*10^-4 mol/hr)*6.022*10^23 ions/mol = 2.563*10^20 ions/hr

(2.563*10^20 ions/hr) * (1/60^2 hr/sec) * (2 elementary charges/ion) * (1.602*10^-19 C/elem charge) = 2.281*10^-2 A

For part (c), 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)*10^5 V = 3.204*10^-14 J/ion

We already calculated that there are

2.563*10^20 ions/hr = 7.118*10^16 ions/sec reaching the collector, so the total energy being dissipated (as heat) in the collector per unit time is:

7.118*10^16 ions/sec * 3.204*10^-14 J/ion = 2.281*10^3 W

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