Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 28, Problem 18P

(a)

To determine

The scattering angle of the photon and the electron.

(a)

Expert Solution
Check Mark

Answer to Problem 18P

The scattering angle of the photon and the electron is mec2+E02mec2+E0_.

Explanation of Solution

The momentum of the photon before scattering is given by,

    p0=hλ0        (I)

Here, p0 is the momentum before scattering, h is the Planck’s constant, λ0 is the wavelength.

The momentum after scattering is given by,

    p=hλ        (II)

The conservation of momentum in the zx direction is given by,

    p0=pcosθ+pecosθ        (III)

Here, pe is the electron momentum after scattering.

Use equation (II) and (I) in equation (III),

    hλ0=(hλ+pe)cosθ        (IV)

The conservation of momentum in the y direction is given by,

    0=psinθpesinθ        (V)

Neglecting the trivial solution will gives,

    pe=p        (VI)

Use equation (VI) in equation (IV),and solve for λ

    hλ0=2hλcosθλ=2λ0cosθ        (VII)

X rays striking a target are scattered at various angles by electrons in the target. A shift in wavelength is observed for the scattered x rays and the phenomenon is known as the Compton effect.

Write the expression for the Compton effect.

    λλ0=hmec(1cosθ)        (VIII)

Use equation (VII) in equation (VIII),

    λλ0=hmec(1cosθ)(2λ0cosθ)λ0=hmec(1cosθ)(2λ0+hmec)cosθ=λ0+hmec(2hmec+hmec)cosθ=hcE0+hcmec        (IX)

Solve equation (IX) for θ.

  1mec2E0(2mec2+E0)cosθ=1mec2E0(mec2+E0)cosθ=mec2+E02mec2+E0        (X)

Conclusion:

Solve equation (IX) for θ.

  1mec2E0(2mec2+E0)cosθ=1mec2E0(mec2+E0)cosθ=mec2+E02mec2+E0        (X)

Therefore, the scattering angle of the photon and the electron is mec2+E02mec2+E0_

(b)

To determine

The energy and the momentum of the scattered photon.

(b)

Expert Solution
Check Mark

Answer to Problem 18P

The energy and the momentum of the scattered photon is 0.602MeV_ and 3.21×1022kgm/s_.

Explanation of Solution

Write the expression for the energy of the scattered photon.

    E=hcλ        (XI)

Write the expression for the momentum after scattering in terms of energy.

    p=Ec        (XII)

Use equation (VII) in equation (XII),

    E=hcλ0(2cosθ)=E02cosθ        (XIII)

Conclusion:

Use equation (X) in equation (XIII),

    E=E02mec2+E02mec2+E0=E0(2mec2+E0)2(mec2+E0)        (XIV)

Use equation (XIV) in equation (XII),

    p=E0(2mec2+E0)2c(mec2+E0)        (XV)

Therefore, The energy and the momentum of the scattered photon is E0(2mec2+E0)2(mec2+E0)_ and E0(2mec2+E0)2c(mec2+E0)_.

(c)

To determine

Kinetic energy and momentum of the scattered electron.

(c)

Expert Solution
Check Mark

Answer to Problem 18P

Kinetic energy and momentum of the scattered electron is 0.278MeV_ and 3.21×1022kgm/s_

Explanation of Solution

Write the expression for the kinetic energy of the electron according to the energy conservation.

    Ke=E0E        (XVI)

Here, Ke is the kinetic energy of the electron.

Use equation (XIV) in equation (XVI),

    Ke=E0E0(2mec2+E0)2(mec2+E0)=2E0(mec2+E0)E0(2mec2+E0)2(mec2+E0)=(2E0mec2+2E02)(2E0mec2+E02)2(mec2+E0)

Solve the above equation,

    Ke=2E0mec2+2E022E0mec2E022(mec2+E0)=E022(mec2+E0)        (XVII)

  pe=p        (XVII)

Conclusion:

Use equation (XV) in equation (XVII) to find pe.

    pe=E0(2mec2+E0)2c(mec2+E0)

Therefore, Kinetic energy and momentum of the scattered electron is E022(mec2+E0)_ and E0(2mec2+E0)2c(mec2+E0)_.

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Chapter 28 Solutions

Principles of Physics: A Calculus-Based Text

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