-3 points HRW9 28 P027 A certain commercial mass spectrometer is used to separat

Question

-3 points HRW9 28 P027 A certain commercial mass spectrometer is used to separate accelerated through a potential difference of 50 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 0.80 m. After traveling My Notes Ask Your uranium ions of mass 3.92 x 10-25 kg and charge 3.20 x 10-19 C from related species. The ions are 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 fleld in the separator? (b) If the machine is used to separate out 160 mg of material per hour, calculate the current of the desired lons in the machine. (c) In that case, calculate the thermal energy produced in the cup in 1.00 h.

Explanation / Answer

Given,

m = 3.92 x 10^-25 kg ; q = 3.2 x 10^-19 kg ; V = 50 kV ; R = 0.8 m ; theta = 180 deg ; w = 1 mm ; h = 1 cm

a)The accelerating potential enables the ions to move, from conservation of energy:

KE = U(electric)

1/2 m v^2 = q V

v = sqrt (2 q V/m)

In spectometer, the magnetic force balances the centipital force:

m v^2/R = q v B

r = mv/Bq = m/Bq sqrt(2 q V/m)

r = 1/B sqrt (2 m V/q)

B = 1/r sqrt (2 m V/q)

B = 1/0.8 sqrt (2 x 3.92 x 10^-25 x 50 x 10^3/(3.2 x 10^-19)) = 0.438 T

Hence, B = 0.438 T

b)We know that

i = nq = (M/m)q

M = 160 x 10^-6/3600 = 4.4 x 10^-8 kg/s

i = (4.4 x 10^-8 / 3.92 x 10^-25) x 3.2 x 10^-19) = 0.036 A

Hence, i = 0.036 A

c)E = n q V t ; but n = i/q

E = (i/q) q V t = i V t

E = 0.036 x 50 x 10^3 x 3600 = 6.48 x 10^6 J

Hence, E = 6.48 x 10^6 J

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