A particle passes through a mass spectrometer as illustrated in the figure below

Question

A particle passes through a mass spectrometer as illustrated in the figure below. The electric field between the plates of the velocity selector has a magnitude of 8390 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.0939 T. In the deflection chamber the particle strikes a photographic plate 11.9 cm removed from its exit point after traveling in a semicircle. What is the mass-to-charge ratio of the particle? What is the mass of the particle if it is doubly ionized? What is its identity, assuming it's an element? (Enter the name of an element.)

Explanation / Answer

a)
Velocity of the particle at the velocity selector can be found out by equating electric force and the magnetic force.
qE = Bqv
v = E/B

Particle is leaving the the particle selector with the velocity E/B and it will now experience only the magnetic force, which will make the particle to move in a circular path.
Any particle moving in circular path experience a centripetal force. Here the magnetic force equals the centripetal force.
Bqv = mv2/r
m/q = Br/v
= rB2/ E
= [0.119 x (0.0939)2 / 8390]
= 12.51 x 10-8

b)
m = q x (m/q)
q = 2e
= 2 x 1.6 x 10-19 C
m = 3.2 x 10-19 C x 12.5059 x 10-8
= 40.032 x 10-27 kg

c)
Mass of hydrogen ion = 1.667 x 10-27 kg = 1u
Mass of the particle in amu = (40.032 x 10-27) / (1.667 x 10-27) = 24 u
This is close to the atomic weight of magnesium

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