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 7830 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.0907 T. In the deflection chamber the particle strikes a photographic plate 39.6 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? (Enter the name of an element.)

Explanation / Answer

The radius of motion = 39.6cm/2 = 19.8cm = 0.198m

For the velocity selector, the electric force balances the magentic force so the particle moves in a straight line:
qE = Bqv
v = E/B = 7830/0.0907 = 86328.55m/s

For the circular motion, the centripetal force is provided by the magnetic force:
mv²/r = Bqv
mv/r = Bq
m/q = Br/v
= 0.0907x0.198/86328.55
= 2.080x10 kg/C
= 2.080x10 kg/C to 3 significant figures.
_________________

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

m = 2.080x10 7 x 3.2x10¹ = 6.656x1026 kg

Since a nucleon has a mass of about 1.67x10²kg and the mass of any electrons is negigible, the particle contains
6.656x10²/(1.67x10²) = 39.85 ~ 40 nucleons

The particle could be one of a number of isotopes of different elements, so asking to identify the element is poor quesiton. After searching through the periodic table one answeris it is a doubly charged calcium ion.

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