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 8572 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.0949 T. In the deflection chamber the particle strikes a photographic plate 39.8 cm removed from its exit point after traveling in a semicircle. What is the mass-to-charge ratio of the particle? kg/C What is the mass of the particle if it is doubly ionized? kg What is its identity, assuming it's an element?

Explanation / Answer

In the velocity selector the electric force and magnetic force are equal

Fe = Fb

E*q = q*v*B

v = E/B = 8572/0.0949 = 90326.6 m/s

in the deflection chamber

distance = s = 2r = 38.6

r = 19.3 cm = 0.193 m

the magnetic force provides the necessary centripetal force

Fb = Fc

q*v*B = m*v^2/r

m/q = rB/v

m/q = (0.193*0.0949)/90326.6

m/q = 2.028*10^-7 kg/C

+++++++++++++++++++

(b)

mass = mass to charge ratio * charge

mass = 2.028*10^-7*2*1.6*10^-19

mass = 6.5*10^-26 kg

++++++++++++++

(c)

potassium

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