An electric field of 160000 V was used to accelerate a singl

An electric field of 16000.0 V was used to accelerate a singly, positively-charged ion with a m/z of 646.2 into the flight tube of a magnetic sector mass spectrometer. What is the velocity of the ion? A homogenous magnetic field with a strength of 4.00 T is applied perpendicular to the length of the flight tube. What is the radius of curvature (in centimeters) for this ion in the magnetic field?

Solution

let the length is L meter

Voltage is 16000 V

So electric field is E= 16000/L V/m

So the force on the particle is, F=zE

So, ma=zE where a is the accelaration

So when the particle will get out the velocity will be,

v^2=u^2+2a*S where v=final velocty, u=initial velocity and S=distance

Here u=0, S=L ,a=zE/m

So,

v^2=2zEL/m=2*16000/646.2 m^2/s^2=49.52 m^2/s^2

v=7.04 m/s

velocity of the ion is 7.04 m/s

b)

magnetic field B=4.0 T

force on the ion.

F=z(v x B)

this sgould be equal to centrifugal force, mv^2/r where r is radius of curvature

zvB=mv^2/r

r=mv^2/(zvB)

=mv/(zB)

=646.2*7.04/4 m

=1137.3 m=113730 cm