An ion that is doubly ionized passes through the velocity selector and into the

Question

An ion that is doubly ionized passes through the velocity selector and into the deflection chamber of a mass spectrometer, as shown below. In the velocity selector the electric field has a magnitude of 8144 V/m, and the magnitude of the magnetic field in both the velocity selector and the deflection chamber is 0.0925 T. If in the deflection chamber the ion is detected at a distance of 11.9 cm from its entry point, determine the following. (a) mass to charge ratio of the ion kg? C (b) mass of the ion kg (c) identity of the ion, assuming it's an element (Use only the masses of elements in their most common form as listed on the periodic table of elements.)

Explanation / Answer

The radius of motion = 11.9 cm/2 = 5.95 cm = 0.0595 m :
qE = Bqv
v = E/B = 8144/0.0925 = 88043.2432 m/s

For the circular motion, the centripetal force is provided by the magnetic force:
mv^2/r = Bqv
mv/r = Bq
m/q = Br/v
= 0.0925x0.0595/88043.2432
= 6.251x10^-8 kg/C

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

m = 6.251x10 ^-8 x 3.2x10^-19 = 2.0x10^-26 kg

c) the particle contains
2.0x10^-26/(1.67x10^-27) = 12 nucleons

doubly charged carbon ion

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