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 multiplied by 10-25 kg and charge 3.20 multiplied by 10-19 C from related species. The ions are accelerated through a potential difference of 110 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 0.80 m. After traveling through 180° and passing through a slit of width 1.00 mm and height 1.00 cm, they are collected in a cup.

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

(b) If the machine is used to separate out 200 mg of material per hour, calculate the current of the desired ions in the machine.


(c) In that case, calculate the thermal energy produced in the cup in 1.00 h.

Explanation / Answer

Conservation of Energy:       0.5 mv2 = qV which equals to, v = v2qV/m            = v2 *( 3.2 x 10-19 ) *( 95 ) * (1000) / (3.92 x10-25 ])            = 39.38 x 104 m/s So,            B = mv / qr               = (3.92 x 10-25 )*( 39.38 x 104) / (3.2 x10-19 * 1.00)               = 0.482 T c) energy:          U = q V             = 3.2 x 10-19 * 95 * 1000             = 3.04 x 10-14 J /ion             = (7.09 x 1016 ion /s) * (3600s /hr) * (3.04 x10-14 J /ion)             = 7759296J

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