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 × 10-25 kg and charge 3.20 × 10-19 C from related species. The ions are accelerated through a potential difference of 101 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 1.41 m. After traveling through 180° and passing through a slit of width 1.23 mm and height 1.39 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 0.921 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 1.25 h.

Explanation / Answer

a) suppose velocity of ion in the region is v then

for circular motion in the magnetic field.

qvB = mv^2/r

B = m v / q r

now for velocity,
KE = q deltaV

m v^2 /2 = q deltV

3.92 x 10^-25 x v^2 /2 = 3.2 x 10^-19 x 101 x 10^3

v = 406076.3 m/s

now putting in values for B,

B = m v / q r = (3.92 x 10^-25 x 406076.3) / (3.20 x 10^-19 x 1.41)

= 0.353 T

b) number of ions in 0.921 mg = 0.921 x 10^-6 / (3.92 x 10^-25)

= 2.35 x 10^12 ions

total charge in 1 hour = 2.35 x 10^12 x 3.20 x 10^-19

current = q/t = (2.35 x 10^12 x 3.20 x 10^-19 C) / (3600 sec)

= 2.09 x 10^-10 A

c) KE of one ion = qdeltav = 3.20 x 10^-19 x 101 x 10^3

thermal energy = number of ion x /ke of one

= 2.35 x 10^12 x 3.20 x 10^-19 x 101 x 10^3
= 0.076 J in 1hour

in 1.25 hr = 0.076 x 1.25 = 0.095 J . Ans

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