Before entering a mass spectrometer, ions pass through a velocity selector consi

Question

Before entering a mass spectrometer, ions pass through a velocity selector consisting of parallel plates that are separated by 2.0 mm and have a potential difference of 160 V. The magnetic field strength is 0.42 T in the region between the plates. The magnetic field strength in the mass spectrometer is 1.2 T. Find (a) the speed of the ions entering the mass spectrometer and the difference in the diameters of the orbits of singly ionized^238U and^235U. The mass of a^235U ion is 3.903 times 10^-25 kg

Explanation / Answer

say v is the velocity of the ions and Q is the charge on the ions then the electric force

             = Eq

Magnetic force = Bvq

for the ions to pass through the two forces must be equal

Bvq = Eq

v = E/B = (160/2.0e-3)0.42   = 1.905 e+5 m/s

b) The radius of the charged particle in a magnetic field is given by

r = mv/BQ

single ionized   q = 1.609e-19 C

                         v = 1.905 e+5 m/s

                         B = 1.2 T

U-235               m = 3.903e-25 kg

U-238               m = 3.903e-25 *238/235 = 3.953e-25

Difference in daimeters of the orbits

                         =2(r(238) -r(2235))

                          = (2v/Bq)(m-238 - m-235)

                          = (2*1.905e+5/1.2*1.609e-19) *0.05e-25

                          = 9.87 e-3 m

                      = 9.87 mm

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