The figure shows a schematic diagram of a simple mass spectrometer, consisting o

Question

The figure shows a schematic diagram of a simple mass spectrometer, consisting of a velocity selector and a particle detector and being used to separate singly ionized atoms (q - +e = 1.602e-19 C) of gold (Au) and molybdenum (Mo). The electric field inside the velocity selector has magnitude E = 1.793-104 V/m and points toward the top of the page, and the magnetic Held has magnitude B1 = 1.37 T and points out of the page. Calculate the velocity, vq, of the ions that make it through the velocity selector (those that travel In a straight line). The gold ions (represented by the black circles) exit the particle detector at a distance d2 = 29.50 cm from the entrance slit, while the molybdenum ions (represented by the gray circles) exit the particle detector at a distance d1 = 14.609 cm from the entrance slit. The mass of a gold Ion Is m gold = 3.27-10-25 kg. Calculate the mass of a molybdenum ion.

Explanation / Answer

Part A)

In the velocity selector, the Electrical Force will be balanced by the Magnetic Force

Therefore F = qE and F = qvB can be set equal

qE = qvB

so v = E/B

v = 1.793 X 104/1.37

v = 1.31 X 104 m/s

Part B)

In the mass separator, the magnetic force causes a centripetal force,

Therefore qvB = mv2/r    (Note, radius is half of are travel distance - convert to meters)

For the gold, we can find B

B = mv/qr

B = (3.27 X 10-25)(1.31 X 104)/[(1.6 X 10-19)(.1475)]

B = .182 T

From that we can find the mass of the molybdenum using the same formula

m = Bqr/v

m = (.182)(1.6 X 10-19)(.073)/(1.31 X 104)

m = 1.623 X 10-25 kg

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