Chapter 39, Problem 20P

(a)

To determine

The scattering angle.

Answer to Problem 20P

The angle $\theta$ is ${\cos }^{-1}\left(\frac{E_0+m{c}^2}{E_0+2m{c}^2}\right)$.

Explanation of Solution

The scattering angle for both the photon and the electron is $\theta$. The energy of the incident photon is $E_0$

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

    $p_x=p^{\prime }\cos \theta +p_e\cos \theta$                                                                                                (I)

Here, $p$ is the momentum of the incident photon in the x-axis, $p^{\prime }$ is the momentum of the scattered photon, $p_e$  is the momentum of the electron and $\theta$ is the scattering angle.

Substitute for $p$ and $p^{\prime }$

  $\frac{h}{{\lambda }_0}=\left(\frac{h}{{\lambda }^{\prime }}+p_e\right)\cos \theta$                                                                                              (II)

Here, $h$ is the Planck’s constant, ${\lambda }_0$ is the wavelength of the incident photon and ${\lambda }^{\prime }$ is the wavelength of the scattered photon.

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

    $0=p^{\prime }\sin \theta -p_e\sin \theta$                                                                                               (III)

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

Rearrange (II) and substitute for $p^{\prime }$

  $p_e=p^{\prime }=\frac{h}{{\lambda }^{\prime }}$                                                                                                         (IV)

Substitute (IV) in (II) and rearrange for ${\lambda }^{\prime }$

  $\begin{array}{l}\frac{h}{{\lambda }_0}=\frac{2h}{{\lambda }^{\prime }}\cos \theta \\ {\lambda }^{\prime }=2{\lambda }_0\cos \theta \end{array}$                                                                                                            (V)

Write the equation for Compton shift

  ${\lambda }^{\prime }-{\lambda }_0=\frac{h}{mc}\left(1-\cos \theta \right)$                                                                                           (VI)

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

Substitute (V) in (VI)

  ${\lambda }_0\left(2\cos \theta -1\right)=\frac{h}{mc}\left(1-\cos \theta \right)$                                                                              (VII)

Rearrange and multiply by $\frac{c}{c}$

  $\left(2\cos \theta -1\right)=\frac{hc}{{\lambda }_{0}m{c}^2}\left(1-\cos \theta \right)$                                                                       (VIII)

Write the expression for $E_0$

  $E_0=\frac{hc}{{\lambda }_0}$                                                                                                                (IX)

Rearrange $\cos \theta$ terms on one side and substitute (IX)

  $\begin{array}{c}\left(2\cos \theta -1\right)+\frac{E_0}{m{c}^2}\cos \theta =\frac{E_0}{m{c}^2}\\ \left(2+\frac{E_0}{m{c}^2}\right)\cos \theta =\frac{E_0}{m{c}^2}+1\\ \cos \theta =\frac{\frac{E_0}{m{c}^2}+1}{\left(2+\frac{E_0}{m{c}^2}\right)}\end{array}$

Rearrange for $\theta$

  $\theta ={\cos }^{-1}\left(\frac{E_0+m{c}^2}{E_0+2m{c}^2}\right)$                                                                              (X)

Thus, the angle $\theta$ is ${\cos }^{-1}\left(\frac{E_0+m{c}^2}{E_0+2m{c}^2}\right)$.

(b)

To determine

The energy and momentum of the photon after scattering.

Answer to Problem 20P

The energy and momentum of the scattered photons are is $\frac{E_0}{2}\left(\frac{E_0+2m{c}^2}{E_0+m{c}^2}\right)$ and $\frac{E_0}{2c}\left(\frac{E_0+2m{c}^2}{E_0+m{c}^2}\right)$.

Explanation of Solution

Substitute for $\cos \theta$ and ${\lambda }_0$ using (X) and (IX) in (V)

  ${\lambda }^{\prime }=\frac{2hc}{E_0}\left(\frac{E_0+m{c}^2}{E_0+2m{c}^2}\right)$                                                                                            (XI)

Write the expression for energy of the scattered photon

  $E^{\prime }=\frac{hc}{{\lambda }^{\prime }}$                                                                                                                (XII)

Here, $E^{\prime }$ is the energy of the scattered photon.

Write the expression for momentum of the scattered photon

  $p^{\prime }=\frac{E^{\prime }}{c}$                                                                                                                (XIII)

Substitute (XI) in (XII)

  $\begin{array}{c}{E}^{\prime }=hc{\left[\frac{2hc}{E_0}\left(\frac{E_0+m{c}^2}{E_0+2m{c}^2}\right)\right]}^{-1}\\ =\frac{E_0}{2}\left(\frac{E_0+2m{c}^2}{E_0+m{c}^2}\right)\end{array}$

Substitute for $E^{\prime }$ in (XIII) to find $p^{\prime }$

    $p^{\prime }=\frac{E_0}{2c}\left(\frac{E_0+2m{c}^2}{E_0+m{c}^2}\right)$

Thus, the energy and momentum of the scattered photons are $\frac{E_0}{2}\left(\frac{E_0+2m{c}^2}{E_0+m{c}^2}\right)$ and $\frac{E_0}{2c}\left(\frac{E_0+2m{c}^2}{E_0+m{c}^2}\right)$.

(c)

To determine

The kinetic energy and momentum of the scattered electron.

Answer to Problem 20P

The kinetic energy and momentum of the scattered photon are $\frac{E_0^2}{2\left(E_0+m{c}^2\right)}$ and $\frac{E_0}{2c}\left(\frac{E_0+2m{c}^2}{E_0+m{c}^2}\right)$ respectively.

Explanation of Solution

Write the expression for kinetic energy of the scattered electron

  $K_{\text{e}}=E_0-E^{\prime }$                                                                                                      (XIV)

Here, $K_{\text{e}}$ is the kinetic energy of the scattered electron.

Substitute for $E^{\prime }$ in (XIV) to find $K_{\text{e}}$

    $\begin{array}{c}{K}_{\text{e}}=E_0-\frac{E_0}{2}\left(\frac{E_0+2m{c}^2}{E_0+m{c}^2}\right)\\ =\frac{2E_0\left(E_0+m{c}^2\right)-E_0\left(E_0+2m{c}^2\right)}{2\left(E_0+m{c}^2\right)}\\ =\frac{E_0^2}{2\left(E_0+m{c}^2\right)}\end{array}$

The momentum of the scattered photon is same as the scattered electron and it is equal to $\frac{E_0}{2c}\left(\frac{E_0+2m{c}^2}{E_0+m{c}^2}\right)$.

Thus, the kinetic energy and momentum of the scattered photon are $\frac{E_0^2}{2\left(E_0+m{c}^2\right)}$ and $\frac{E_0}{2c}\left(\frac{E_0+2m{c}^2}{E_0+m{c}^2}\right)$.