Question

It is then deflected by a uniform magnetic field (perpendicular to the ion's velocity) into a semicircle of radius R. Now a triply charged ion of mass m' isaccelerated through the same potential difference and deflected by the same magnetic field into a semicircle of radius R' = 4R. What is the ratio of the masses of theions? m'/m =

Answer 1

First we notice that when an ion is accelerated through acertain potential difference V it will reach a certain velocity v.Then once we find thevelocity we can determine what is themagnetic force the ion experiences when it enters the magneticfield. We also know that the magnetic force willmake the particlerotate about a circle and that when the particle rotates throughthe circle it will experience a centripital force.The Velocity vcan be determinedknowingthat the change in kinetic energy isequal to thecharge times thepotentialdifference.ΔKE= qVsince the firstion starts at restΔKE =(1/2) m1v1^2 where v1 is the velocity, q1 thecharge,m1 the mass of thefirst iontherefore (1/2) m*v1^2=q1*V you can solve for the velocity v. v1 = √[(2*q1*V) / (m1)]similarly solving for the velocity ofthe second ion we get v2 =√[ (2*q2*V) / (m2)]where v2 is the velocity, q2 the charge,m2the mass of thefirst ion

we know that q2 = 3 * q1 replacing q2intermsof q1 we getv2 = √[ (2*(3q1)*V) /(m2)] =√[ (6*q1*V) / (m2)]


sofar wev1 =√[ (2*q1*V) / (m1)] andv2 = √[ (6*q1*V) /(m2)]

we know that the magnetic forceFmion 1will experiece is Fm = (q1 * v1)x Band since thevelocityand the magnetic field areperpendicular Fm =q1*v1*B

We also know that the centripital force Fc ion 1will experience is Fc= (m1 * v1^2) /R

We alsoknow that because Fm = Fcso(q1 * v1)x B =(m1 * v1^2)/R. Then solve form1

you get m1 = (q1*R*B)/v1similarly solving for m2 = (q2 * 4R * B)/ v2

Now to finish divide m2/m1after some cancellationof variable m2/m1= (12v1/v2).Then substitute thevalueswe obtained for v1and v2 and solve further.

and you should get m2/m1= 48

Hopefully, this helped. I am new toCramster so i don'tknowhow to make the equation look nice.