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 7778 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.0904 T. In the deflection chamber the particle strikes a photographic plate 58.3 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

here,
A.)
radius = 58.3 /2 = 29.3 cm = 0.293 m

In a Velocity selector Electric field balance with Magnatic Field so parrticle can move in staright line,ThereFore
qE = Bqv

V = E /B
V = 7778 / 0.0904
V = 86039.82 m/s
From Circular Motin Centripital force will be provided by Manatic Force Therefore We get,

MV^2/R = B*q*V
M/Q = BR/v
M/Q = ( 0.0904 * 0.293 ) / 86039.82
M/Q = 3.07 *10^-7 Kg/C

The Mass to Charge ratio is 3.07 *10^-7 Kg/C

B)
In case of doubled ionised charge
Q = charge of Electron * 2
Q = 2x1.6*10^-19
Q = 3.2x10^-19 C

ThereFore mass will be Equal to
M/Q = 3.07 *10^-7 Kg/C
M = 3.07 *10^-7 * 3.2x10^-19
M = 9.824 *10^-26 Kg

mass in case of doubled ionised charge is 9.824 *10^-26 Kg

C.)
Mass of Nucleon = 1.67*10^-27 kg
mass of Electron is Negligible as compared to mass of Nucleon,so all mass will be from protons and Neutrons

No of Nucleons = 9.824 *10^-26 / 1.67*10^-27
n = 58.82 = 59

The Doubled Charge ion is isotope of Cobalt which have 27 Protons and 32 Neutrons.

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