Question

A certain commercial mass spectrometer is used to separate uranium ions of mass 3.92 × 10-25 kg and charge 3.20 × 10-19 C from related species. The ions are accelerated through a potential difference of 106 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 0.891 m. After traveling through 180° and passing through a slit of width 0.774 mm and height 1.45 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 1.23 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 1.32 h.

Answer 1

Given,

m = 3.92 x 10^-25 kg

e = 3.2 x 10^-19 C

V = 106 x 10³ V

R = 0.891 m

w = 0.774 x 10^-3 m

h = 1.45 x 10^-2 m

A)

A particle which is moving with velocity v and perpendicular to the magnetic field then It move in a circular path,

Also, a charged particle accelerated by a voltage V having kinetic energy

Put the value of v in the equation of magnetic field, we have

Putting all values we have then

B = 0.5719 T

B)here given,

1.23 mg/hr = 0.00123 gm/hr

Now,

0.00123 (gm/hr) ÷ 235 (gm/mol) = 5.23 x 10^-6 mol/hr

Again convert into ions,

5.23 x 10^-6 (mol/hr) x 6.022 x 10²³ (ions/mol) = 3.1519 x 10¹⁸ ions/hr

Finally,

3.1519 x 10¹⁸ (ions/hr) x (1÷ 60² hr/sec) x 2 (elementary charge /ionions) x (1.6 x 10^-19) C/elementray charge

= 2.8 x 10^-4 A.

C)

The kinetic energy for each ions is taken out by

= eV

= 2 x 1.6 x 10^-19 x 106 x 10³

= 339.2 x 10^{-16 J/ions}

We have already calculated that there are

3.1519 x 10¹⁸ ions / hr

Convert into second by dividing the above number by 3600

= 0.087 x 10¹⁶ ions / sec

Therefore,

= 0.087 x 10¹⁶ (ions/sec) x 339.2 x 10^-16 (J/ions)

= 29.68 J/sec

Multiply the above answer by converting 1.32 hr into second for part C answer