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 0.50 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 110 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

Given that The mass of uranium ions, m = 3.92 x10^-25 kg charge of uranium ions, q = 3.2 x10^-19 C The potential difference through ions are accelerating, v = 120 kv = 110x10^3 kg a) From the law of conservation of energy (1/2)mv^2 = qV v = v(2qV/m) = v[2( 3.2 x10^-19 C)(110x10^3 kg )/( 3.92 x10^-25 kg)] = 4.24 x10^5 m/s B = mv / qr = (3.92 x 10^-25 kg )( 4.24x10^5 m/s) / (3.2 x10^-19C) (0.5 m)) = 1.0388 T b) Current is I = qM/m =( 3.2x 10^-19C) (110x 10^-6 kg/(3600 s))/(3.92 x10^-25 kg) =0.0249 A c) Thermal energy E = I ?V t = (0.0249 A)(110 x 10^3 V)(3600s) =9.56 x10^7 J

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