The electric field between the plates of the velocity selector in a Bainbridge m

Question

The electric field between the plates of the velocity selector in a Bainbridge mass spectrometer (see Fig. 27.22 in the textbook) is 1.29×105 V/m , and the magnetic field in both regions is 0.550 T . A stream of singly charged selenium ions moves in a circular path with a radius of 35.0 cm in the magnetic field.

Determine the mass of one selenium ion.

Determine the mass number of this selenium isotope. (The mass number is equal to the mass of the isotope in atomic mass units, rounded to the nearest integer. One atomic mass 1u=1.66×1027kg.)

Explanation / Answer

E = 1.29*10^5 V/m , B =0.55 T, r = 35 cm

Electric force = magnetic force

qE = qvB

1.29*10^5 = v*0.55

V = 2.345*10^5 m/s

q =1.6*10^-19 C

Magnetic force = cnetripetal force

qvB =mv^2/r

qB = mv/r

1.6*10^-19*0.55 = m*2.345*10^5/0.35

m =1.313*10^-25 kg

Atomic mass M = m/1u = (1.313*10^-25)/(1.66*10^-27)

M = 79

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