Question

In a mass spectrometer, a student needs to find the strength of the electric field that is needed in the velocity sector if a single charged particle with a mass of 7.971x10^-26 kg after leaving the velocity sector is to hit the ceiling 0.036 m above the opening where it enters the region with only magnetic field. There is no E field. The value of the magnetic field is 0.025 T in the entire mass spec.
A) what is the speed or velocity of the particles as they leave the velocity sector and what is the electric fields strength needed in the velocity sector ?

B) if there is a particle with a charge of -2e and a mass of 5.314x10^-26 kg, where will it end up in this mass spec if it makes it through the velocity sector? Draw the path of the new particle and calculate where the particle will circulate back to hit the ceiling

3.

Answer 1

Mass of the particle in the mass spectrometer=7.971*10^-26 kg

Distance of the ceiling from the opening r=0.036 m

Magnetic field in the region=0.025 T

a) When the particle emerges out of the opening, it follows a parabolic path. Hence a centripetal force appears.

This is given by

F_c=mv²/r

where v=velocity of the field as it leaves the velocity sector

The magnetic force is given by F_B=qvB

where q=charge of the particle(electron)=1.6*10^-19C

Equating the two forces, v=qBr/m=1.6*10^-19*0.025*0.036/(7.971*10^-26)=1.81*10³m/s

The electric field strength needed in this velocity sector E=vB=1.81*10³*0.025=45 V/m

b) If the particle has a charge of q=-2e or -3.8*10^-19C, and mass of m= 5.314*10^-26 kg,

Equating qvB=mv²/r

r=mv/qB=5.314*10^-26*1.81*10³/(3.8*10^-19*0.025)=101.25*10^-4m=0.0101 m

This is the end point of the particle in the velocity vector of the mass spectrometer.