Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 39, Problem 20P

(a)

To determine

The scattering angle.

(a)

Expert Solution
Check Mark

Answer to Problem 20P

The angle θ is cos1(E0+mc2E0+2mc2).

Explanation of Solution

The scattering angle for both the photon and the electron is θ. The energy of the incident photon is E0

Write the expression for conservation of momentum in the x-direction

    px=pcosθ+pecosθ                                                                                                (I)

Here, p is the momentum of the incident photon in the x-axis, p is the momentum of the scattered photon, pe  is the momentum of the electron and θ is the scattering angle.

Substitute for p and p

  hλ0=(hλ+pe)cosθ                                                                                              (II)

Here, h is the Planck’s constant, λ0 is the wavelength of the incident photon and λ is the wavelength of the scattered photon.

Write the expression for conservation of momentum in the y-direction

    0=psinθpesinθ                                                                                               (III)

Here, py is the momentum of the incident photon in the y-axis.

Rearrange (II) and substitute for p

  pe=p=hλ                                                                                                         (IV)

Substitute (IV) in (II) and rearrange for λ

  hλ0=2hλcosθλ=2λ0cosθ                                                                                                            (V)

Write the equation for Compton shift

  λλ0=hmc(1cosθ)                                                                                           (VI)

Here, m is the mass of electron and c is the speed of light.

Substitute (V) in (VI)

  λ0(2cosθ1)=hmc(1cosθ)                                                                              (VII)

Rearrange and multiply by cc

  (2cosθ1)=hcλ0mc2(1cosθ)                                                                       (VIII)

Write the expression for E0

  E0=hcλ0                                                                                                                (IX)

Rearrange cosθ terms on one side and substitute (IX)

  (2cosθ1)+E0mc2cosθ=E0mc2(2+E0mc2)cosθ=E0mc2+1cosθ=E0mc2+1(2+E0mc2)

Rearrange for θ

  θ=cos1(E0+mc2E0+2mc2)                                                                              (X)

Thus, the angle θ is cos1(E0+mc2E0+2mc2).

(b)

To determine

The energy and momentum of the photon after scattering.

(b)

Expert Solution
Check Mark

Answer to Problem 20P

The energy and momentum of the scattered photons are is E02(E0+2mc2E0+mc2) and E02c(E0+2mc2E0+mc2).

Explanation of Solution

Substitute for cosθ and λ0 using (X) and (IX) in (V)

  λ=2hcE0(E0+mc2E0+2mc2)                                                                                            (XI)

Write the expression for energy of the scattered photon

  E=hcλ                                                                                                                (XII)

Here, E is the energy of the scattered photon.

Write the expression for momentum of the scattered photon

  p=Ec                                                                                                                (XIII)

Substitute (XI) in (XII)

  E=hc[2hcE0(E0+mc2E0+2mc2)]1=E02(E0+2mc2E0+mc2)

Substitute for E in (XIII) to find p

    p=E02c(E0+2mc2E0+mc2)

Thus, the energy and momentum of the scattered photons are E02(E0+2mc2E0+mc2) and E02c(E0+2mc2E0+mc2).

(c)

To determine

The kinetic energy and momentum of the scattered electron.

(c)

Expert Solution
Check Mark

Answer to Problem 20P

The kinetic energy and momentum of the scattered photon are E022(E0+mc2) and E02c(E0+2mc2E0+mc2) respectively.

Explanation of Solution

Write the expression for kinetic energy of the scattered electron

  Ke=E0E                                                                                                      (XIV)

Here, Ke is the kinetic energy of the scattered electron.

Substitute for E in (XIV) to find Ke

    Ke=E0E02(E0+2mc2E0+mc2)=2E0(E0+mc2)E0(E0+2mc2)2(E0+mc2)=E022(E0+mc2)

The momentum of the scattered photon is same as the scattered electron and it is equal to E02c(E0+2mc2E0+mc2).

Thus, the kinetic energy and momentum of the scattered photon are E022(E0+mc2) and E02c(E0+2mc2E0+mc2).

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