Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 29, Problem 63AP
To determine

The radius of the alpha particle’s trajectory.

Expert Solution & Answer
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Answer to Problem 63AP

The radius of the alpha particle’s trajectory is 3R4.

Explanation of Solution

Write the equation for the radius of the trajectory of the proton.

    R=mpv1Bq1                                                                                                                   (I)

Here, R is the radius of the trajectory of the proton, mp is the mass of the proton, v1 is the velocity of the proton after the collision, B is the magnetic field and q1 is the charge on the proton.

Write the equation for the radius of the trajectory of the alpha particle.

    R=mav2Bq2                                                                                                                 (II)

Here, R is the radius of the trajectory of the alpha particle, ma is the mass of the alpha particle, v2 is the velocity of the alpha particle after the collision and q2 is the charge on the alpha particle.

Write the equation for the principle of conservation of linear momentum.

    mav=mpv1+mav2                                                                                                    (III)

The alpha particle's mass is four times the proton's mass.

    ma=4mp

Substitute 4mp for ma in the equation (III).

    4mpv=mpv1+4mpv24v=v1+4v2                                                                                                (IV)

Write the equation for the principle of conservation of energy.

    12mav2=12mpv12+12mav22                                                                                        (V)

Substitute 4mp for ma in the above equation.

    12(4mp)v2=12mpv12+12(4mp)v224v2=v12+4v22                                                                            (VI)

Take the square of equation (IV) and then divide by 4.

    (4v)24=(v1+4v2)244v2=(v1+4v2)24                                                                                                (VII)

Compare the equation (VI) and (VII).

    v12+4v22=(v1+4v2)244v12+16v22=v12+16v22+8v1v23v12=8v1v23v1=8v2                                                                               (VIII)

Rewrite the equation (I).

    v1=Bq1Rmp

Rewrite the equation (II).

    v2=Bq2Rma

Substitute Bq1Rmp for v1 and Bq2Rma for v2 in the equation (VIII).

    3(Bq1Rmp)=8(Bq2Rma)

The alpha particle's charge is twice the proton's charge.

    q2=2q1

Substitute 2q1 for q2 and 4mp for ma in the above equation.

    3(Bq1Rmp)=8(B(2q1)R4mp)3R=4RR=3R4

Conclusion:

Therefore, the radius of the alpha particle’s trajectory is 3R4.

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Chapter 29 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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