A proton is at rest at the plane vertical boundary of a region containing a unif

Question

A proton is at rest at the plane vertical boundary of a region containing a uniform vertical magnetic field B. An alpha particle moving horizontally makes a head-on elastic collision with the proton. Immediately after the collision, both particles enter the magnetic field, moving perpendicular to the direction of the field. The radius of the proton's trajectory is 2.5 R. Find the radius of the alpha particle's trajectory as a factor of R. The mass of the alpha particle is four times that of the proton, and its charge is twice that of the proton.
r = R

Explanation / Answer

after collision let velocity of proton = Vp and alpha particle be = Va also mass of the alpha particle is four times that of the proton Ma = 4*Mp

Va = [(Ma-Mp)/(Ma+Mp)}*V = 3/5*V

Vp = [2Ma /(Ma+Mp)]*V = 8/5*V

radius of a charged particle in magnetic field is given by

r=M*V/QB

where Q= charge , B=magnetic field

given The radius of the proton's trajectory is 2.5 R

2.5R = Mp*Vp/QpB = Mp*(8/5V)/QpB  

hence the radius of the alpha particle trajectory be R'

R' = MaVa/QaB = 4Mp*(3/5)V/2QpB = 1.875R

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