A mass spectrometer, sketched below, is a device used to separate different ions

Question

A mass spectrometer, sketched below, is a device used to separate different ions. Such ions with a well-defined velocity v_o enter through a slit into a region of uniform magnetic field B, where they follow a semicircular path until they strike the detector slit above the entry slit. The distance between the entry and the detector slits is d = 1.30 m. Chlorine ions of mass 35 amu (1 amu equals 1.66 Times 10^-27 kg), carrying a charge of +1e enter the spectrometer with initial speed of v_o = 7.60 Times 10^5 m/s. What value of B is required for these ions to hit the detector? Equally charged chlorine ions of mass 37 amu now enter the spectrometer. How close to the detector slit will they impact?

Explanation / Answer

(a)

d = 2 R
Where R is the radius of the semi circle.
R = d/2 = 1.3m/2 = 0.65m
Mass of Particle = 35amu = 35 * 1.66*10^-27 Kg

Force on Charge moving with velocity v in Magnetic Field B is given by . F = q*v*B

Magnetic force serves as the centripetal force, we have:
m*v^2/ R = q v B
B = (m*v^2/R) / (q*v)
B = (35 * 1.66*10^-27 * 7.6*10^5)/(1.6*10^-19 * 0.65 )
B = 0.425 T

(b)
Mass = 37amu = 37 * 1.66*10^-27 Kg

m*v^2 / R = q v B
R = m*v^2/q v B
R = m*v/q*B
R = (37*1.66*10^-27 * 7.6*10^5)/(1.6*10^-19 * 0.425)
R = 0.686

d = 2*R
d = 2*0.686
d = 1.372 m

Distance from the detector slit = 1.372 - 1.3
Distance from the detector slit = 0.072m

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