Question
1. The diagram shows a two-chamber mass spectrometer device. Singly-charged ions (q = +|e|)
enter from the left. They first pass through the
The diagram shows a two-chamber mass spectrometer device. Singly-charged ions (q = +|e|) enter from the left. They first pass through the "velocity selector", a chamber of length L=1.99 meters where there is a magnetic field B = 0.48 Tesla pointing into the page and a vertical electric field E. The E field is created by a voltage V applied across the upper and lower plates shown (of separation d=0.39 meters). The E and B fields are chosen such that only particles of a certain velocity v pass undeflected through the tiny hole at the end of the chamber. The ions have net charge q = +|e| and mass m. If we want to select only those with velocity v=84 km/sec, what voltage V (in kilovolts) must we apply across the plates of the velocity selector? After exiting the velocity selector, the ions enter the actual mass spectrometer. In this region the same B field exists (B=0.19 T, directed into the paper), but there is no electric field. The ions curve upward, and are then detected. By measuring the diameter of the trajectory, the mass of the singly-charged ions (q=+|e|=1.6x10-19 Coulomb) can be determined. If we want 6Li (mass = 1.00 x 10-26 kg) and 7Li (mass = 1.17 x 10-26 kg) to be separated by a distance x = 0.17 cm at the detector (see figure), what must be the velocity v of the ions entering the mass spectrometer?
Explanation / Answer
1) Velocity selector
for a certain velocity of pass undeflected
magnetic force-electric force
==> qvB=qE
==> vB=E=V/d
==> V=dvB= 0.39*84000*0.48= 15724.8 Volts
2) spectrometer region
Radius of the path==>
mv^2/R=qvB
==> R= mv/qB
R1=m1v/qB and R2=m2v/qB
==> X=2*(R2-R1)= 2*(m2-m1)*v/qB
==> 0.0017 = 2*(1.17E-26-1E-26)*v/(1.6E-19*0.19)
==> v= 0.0017*(1.6E-19*0.19)/2*(1.17E-26-1E-26) = 15200 m/s
