Question

Consider the mass spectrometer shown schematically in the figure below. The magnitude of the electric field between the plates of the velocity selector is 2.60  10³ V/m, and the magnetic field in both the velocity selector and the deflection chamber has a magnitude of 0.0300 T. Calculate the radius of the path for a singly charged ion having a mass

m = 2.48  10^-26 kg

Answer 1

by force balance on singly charged ion in velocity selector

electric force = magnetic force

Fe = Fm

q*E = q*V*B

V = E/B

Now when ions passes through magnetic field in the deflection chamber, radius of their paths will be

Using force balance on charged particle

Centripetal force = Magnetic force

Fc = Fm

m*V^2/R = q*V*B

r = m*V/(q*B)

r = m*E/(q*B^2)

given , E = 2.60*10^3 V/m

B = 0.0300 T

q = single charged = 1.6*10^-19 C

m = mass of charged ion = 2.48*10^-26 kg

So,

r = 2.48*10^-26*2.60*10^3/(1.6*10^-19*0.0300^2)

r = 0.4478 m = 44.78 cm = radius of the path followed by charged ion

Let me know if you've any query.