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 8021 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.0918 T. In the deflection chamber the particle strikes a photographic plate 58.3 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?

" x xxxx 10[s APA x xxx xxx xxx xx x x -- -3 x xxx xx xx x xx xx xx x x xx xxx xx xx xxx xxx xxx xxxx x. 4 plate Photographic

Explanation / Answer

given data

E = 8021 V/m

B = 0.0918 T

dimater of the path follwed by the charged particle, d = 58.3 cm = 0.593 m

radius of the path, r = d/2 = 0.583/2 = 0.2915 m

a) In velovity selector,

Fe = FB

q*E = q*v*B

==> v = E/B

= 8021/0.0918

= 8.74*10^4 m/s

we know, r = m*v/(B*q)

m/q = B*r/v

= 0.0918*0.2915/(8.74*10^4)

= 3.06*10^-7 kg/C

b) m = 3.06*10^-7*q

= 3.06*10^-7*2*1.6*10^-19

= 9.79*10^-26 kg

c) mass number of element, A = 9.79*10^-26/(1.67*10^-27)

= 57

so the element is La(Lanthumum)

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