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 8091 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.0922 T. In the deflection chamber the particle strikes a photographic plate 11.9 cm removed from its exit point after traveling in a semicircle.

(a) What is the mass-to-charge ratio of the particle?
kg/C

(b) What is the mass of the particle if it is doubly ionized?
kg

(c) What is its identity, assuming it's an element?

Explanation / Answer

The radius of motion = 11.9/2 cm = 5.95 cm = 0.0595m

For the velocity selector, the electric force balances the magentic force so the particle moves in a straight line:
qE = Bqv
v = E/B = 8091/0.0922 = 87754.88 m/s

For the circular motion, the centripetal force is provided by the magnetic force:
mv²/r = Bqv
mv/r = Bq
m/q = Br/v
= 0.0922x0.0595/87754.88
= 6.251x108 kg/C
= 6.25x108 kg/C to 3 significant figures.
_________________

If it is doubly ionised its charge = 2x1.6x10¹ C = 3.2x10¹C

m = 6.251x108 x 3.2x10¹ = 2.00x10² kg

Since a nucleon has a mass of about 1.67x10²kg and the mass of any electrons is negigible, the particle contains
2.00x10²/(1.67x10²) = 11.98 = 12 nucleons

After searching through the periodic table one answeris it is a doubly charged carbon ion.

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