A particle passes through a mass spectrometer as illustrated in the figure below
ID: 2056509 • Letter: A
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.(a) What is the mass-to-charge ratio of the particle?
.
Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. 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?
Explanation / Answer
If we consider a Mass spectrometry experiment, In that a charged particle is subjected to an electric and magnetic field, causing the particles to separate on the basis of their charge and mass ratio.
The magnitude of the deflection of the moving ion's trajectory depends on its mass-to-charge ratio. FInally, one can measure the radius in which a particle hit the detection plate.
Its given by the equation :
q v B = (m v2)/r (v = Velocity of Particle, B = Magnetic Field Strength, m = mass of particle, r = radius of particle's movement)
m / q = B r / v - (1)
Now, 0.5 m v2 = q V (Where V is Voltage) ( By Energy conservation) - (2)
solving ( 1) and (2), we get
v = 2V/Br (All given in question, r = 39.8/2 = 19. 9 cm , B = 0.0949 T)
v = 909805.62 ms-1
Now, m/q = 2.07 x 10-8
(b) Mass of Partcle when charge is 2e
m = 2 x 1.6 x 10-19 x 2.07 x 10-8 = 6.64 x 10-27 kg
(c) It has approximately 4 times the mass of Hydrogen or 4 amu ( 1 atomic mass units = 1.66053886 × 10-27 kilograms)
Hence, the element is He (4 amu)
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