Question

he electric field between the electric plates in the mass spectrometer is 2.58

Answer 1

please follow concept you will get answer

In a mass-spec. the ions that get out have velocity "v" determined by the E & B fields as;

qE = qvB

v = E/B

Once out of the slit, the ions will be forced into circular motion of radius "R" by the magnetic force qvB' , and this equals the centripetal force;

qvB' = mv^2/R

The radius is then;

R = mv/qB' = mE/qBB'

The ions will travel in a semicircle and hit the plate a distance 2R from the exit slit. So for each mass m1, m2, & m3 the distance 2R will be slightly different.

2R1 = 2m1E/qBB'

2R2 = 2m2E/qBB'

2R3 = 2m3E/qBB'

The spacing are then;

D12 = 2R2 - 2R1

D23 = 2R3 - 2R2

When the ions are doubly sharged just replace "q" by "2q" , where q = 1.6x10^-19 C