PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 26, Problem 51P

(a)

To determine

The speed of the electrons produced by the electron accelerator.

(a)

Expert Solution
Check Mark

Answer to Problem 51P

The speed of the electrons produced by the electron accelerator is 0.99980c_.

Explanation of Solution

Given that the kinetic energy of the electrons is 25MeV, the rest mass energy of electron is 0.5110MeV.

Write the expression for relativistic kinetic energy.

    K=(γ1)mc2                                                                                                         (I)

Here, K is the kinetic energy of the electrons, γ is the Lorentz factor, m is the mass of the particle, c is the speed of light.

Write the expression for the Lorentz factor.

    γ=11v2/c2                                                                                                        (II)

Here, v is the speed of the spaceship.

Use equation (II) in equation (I).

    K=(11v2c21)mc2                                                                                            (III)

Solve equation (III) for v.

    1+Kmc2=11v2c21v2c2=(1+Kmc2)2v=c1(1+Kmc2)2                                                                                  (IV)

Conclusion:

Substitute 25MeV for K, 0.5110MeV for mc2 in equation (IV) to find v.

    v=c1(1+25MeV0.5110MeV)2=0.99980c

Therefore, the speed of the electrons produced by an 0.99980c_.

(b)

To determine

The time taken for the electrons to travel the distance in the reference frame of electrons.

(b)

Expert Solution
Check Mark

Answer to Problem 51P

The time taken for the electrons to travel the distance in the reference frame of electrons is 0.010ns_.

Explanation of Solution

Write the expression for the time taken in the reference frame of electrons.

    Δt0=Δtγ                                                                                                                 (V)

Here, Δt0 is the time interval in the reference frame of the electrons, Δt is the time taken by the electrons to exit the accelerator and reach the patient in the reference frame.

Write the expression for the time taken by the electrons to exit the accelerator and reach the patient in the reference frame.

    Δt=Δxv                                                                                                                (VI)

Here, Δx is the distance of the accelerator from the patient, v is the speed of the electrons.

Use equation (II) and (VI) in equation (V),

    Δt0=Δx1v2c2v                                                                                                  (VII)

Conclusion:

Substitute 15cm for Δx, 0.99980c for v, and 2.998×108m/s for c in equation (VII) to find Δt0.

    Δt0=15.0cm×1(0.99980c)2c20.99980(2.998×108m/s)=(15.0cm×1m100cm)1(0.99980c)2c20.99980(2.998×108m/s)=0.010ns

Therefore, the time interval in the reference frame of electrons is 0.010ns_.

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Chapter 26 Solutions

PHYSICS

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