A certain commercial mass spectrometer (Fig. 28-12) is used to separate uranium

Question

A certain commercial mass spectrometer (Fig. 28-12) is used to separate uranium ions of mass 3.92 x 10-25 kg and charge 3.20 x 10-19 C from related species. The ions are accelerated through a potential difference of 99.5 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 1.35 m. After traveling through 180° and passing through a slit of width 1.34 mm and height 1.22 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 1.32 mg of material per hour, calculate (b) the current (in A) of the desired ions in the machine and (c) the thermal energy (in J) produced in the cup in 1.29 h

Explanation / Answer

Here the total energy of the ions is conserved
Therefore
K+U = 0
0.5*mv2 - qV = 0

v = (2qV/m)

The radius of the circular path taken by the ions is given by

r = mv/Bq

r = m/Bq((2qV/m))

As x = 2r

m = B2qx2/8V

B = (8Vm/qx2)

B = [(8*99.5*103 V*3.92 x 10-25 kg)/(3.20 x 10-19 C*2.7*2.7 m2)]

B = 0.134 T

b)

Let N be the number of ions that are separated by the machine per unit time. The
current is i = qN and the mass that is separated per unit time is M = mN, where m is the
mass of a single ion. M has the value

M = 1.32*10-6 kg / 3600 s

M = 3.67*10-10kg/s

Since N= M/m

I = qM/m

I = (3.20 x 10-19 C)(3.67*10-10kg/s)/(3.92 x 10-25 kg)

I = 2.99*10-4A

c)

Each ion deposits energy qV in the cup, so the energy deposited in time t is given by

E = NqVt

E = IVt

E = ( 2.99*10-4A)(99.5*103 V)(1.29*3600 s)

E = 1.38*105 J

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