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Chapter 28, Problem 17P

(a)

To determine

The scattering angle of the photon and the electron.

(a)

Expert Solution
Check Mark

Answer to Problem 17P

The scattering angle of the photon and the electron is 43.0°_

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:

Substitute 0.511MeV for mec2 , 0.880MeV for E0 in equation (X) to find θ .

  cosθ=0.511MeV+0.880MeV2(0.511MeV)+0.880MeV=0.731θ=cos1(0.731)=43.0°

Therefore, The scattering angle of the photon and the electron is 43.0°_

(b)

To determine

The energy and the momentum of the scattered photon.

(b)

Expert Solution
Check Mark

Answer to Problem 17P

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:

Substitute 0.880MeV for E0 and 43.0° for θ in equation (XIII) to find E.

    E=0.880MeV2cos43.0°=0.602MeV

Substitute 0.602MeV for 3×108m/s for c in equation (XII) to find p.

    p=0.602MeV3×108m/s=3.21×1022kgm/s

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

(c)

To determine

Kinetic energy and momentum of the scattered electron.

(c)

Expert Solution
Check Mark

Answer to Problem 17P

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        (XIV)

Here, Ke is the kinetic energy of the electron,

  pe=p        (XV)

Conclusion:

Substitute 0.880MeV for E0 and 0.602MeV for E in equation (XIV) to find Ke.

    Ke=0.880MeV0.602MeV=0.278MeV

Substitute 3.21×1022kgm/s for p to find pe.

    pe=3.21×1022kgm/s

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

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Chapter 28 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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