   Chapter 19, Problem 19P

Chapter
Section
Textbook Problem

A proton is at rest at the plane vertical boundary of a region containing a uniform vertical magnetic field B (Fig. P19.19). 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 R. The mass of the alpha particle is four times that of the proton, and its charge is twice that of the proton. Find the radius of the alpha particle’s trajectory. Figure P19.19

To determine
The radius of the alpha particles trajectory.

Explanation

Given info: Initially the proton is stationary. The proton and the alpha particle undergo elastic collision. Both the particles move perpendicular to the uniform magnetic field. The radius of the protons trajectory is R , Initially the proton is stationary. The proton and the alpha particle undergo elastic collision. Both the particles move perpendicular to the uniform magnetic field. The radius of the protons trajectory is R .

Explanation:

The mass of the alpha particle is four time that of the proton,

mαmp=4       (1)

• mα is the mass of the alpha particle
• mp is the mass of the proton

The charge of alpha particle is twice of proton,

qαqp=2       (2)

• qα is the charge of the alpha particle
• qp is the charge of the proton

Since the collision is elastic, the linear momentum as well as the kinetic energy in the collision is conserved.

From conservation of linear momentum,

mαvo=mαvα+mpvp       (3)

• vo is the initial velocity of the alpha particle

Re-arrange to get vo ,

vo=mαvα+mpvpmα       (4)

From conservation of kinetic energy,

mαvo2=mαvα2+mpvp2       (5)

Combining (4) and (5),

mα(mαvα+mpvpmα)2=mαvα2+mpvp2mα2vα2+mp2vp2+2mαvαmpvp=mα2vα2+mαmpvp22mαvα=vp(mαmp)vαvp=mαmp2mα

Since mα=4mp ,

vαvp=4mpmp8mp=38

Hence the ratio of the velocities of the alpha particle and the proton is 38

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