Chapter 40, Problem 31P

(a)

To determine

The scattering angle of the photon and the electron.

Answer to Problem 31P

The scattering angle of the photon and the electron is $43.0°$ .

Explanation of Solution

Write the equation for the momentum of the photon before scattering.

  $p_0=\frac{E_0}{c}$                                                                                                                     (I)

Here, $p_0$ is the momentum of the photon before scattering, $E_0$ is the energy of the photon before scattering and $c$ is the speed of light in vacuum.

Write the equation for the energy of the incident photon.

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

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

Put equation (II) in equation (I).

  $\begin{array}{c}{p}_0=\frac{hc/{\lambda }_0}{c}\\ =\frac{h}{{\lambda }_0}\end{array}$                                                                                                              (III)

Write the equation for the momentum of the photon after scattering.

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

Here, $p^{\prime }$ is the momentum of the photon after scattering and $E^{\prime }$ is the energy of the photon after scattering.

Write the equation for the energy of the scattered photon.

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

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

Put equation (V) in equation (IV).

  $\begin{array}{c}{p}^{\prime }=\frac{hc/{\lambda }^{\prime }}{c}\\ =\frac{h}{{\lambda }^{\prime }}\end{array}$                                                                                                             (VI)

Refer to figure P40.31,and write the equation for the conservation of momentum in $x$ direction.

  $p_0=p^{\prime }\cos \theta +p_e\cos \theta$

Here, $\theta$ is the angle of scattering and $p_e$ is the momentum of the electron after the scattering.

Put equations (III) and (VI) in the above equation.

  $\begin{array}{c}\frac{h}{{\lambda }_0}=\frac{h}{{\lambda }^{\prime }}\cos \theta +p_e\cos \theta \\ =\left(\frac{h}{{\lambda }^{\prime }}+p_e\right)\cos \theta \end{array}$                                                                                         (VII)

Refer to figure P40.31,and write the equation for the conservation of momentum in $y$ direction.

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

Neglect the trivial solution $\theta =0$ for the above equation.

  $\begin{array}{c}{p}^{\prime }\sin \theta =p_e\sin \theta \\ p^{\prime }=p_e\end{array}$                                                                                                (VIII)

Put equation (VI) in the above equation.

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

Put equation (IX) in equation (VII) and rewrite it for ${\lambda }^{\prime }$ .

  $\begin{array}{c}\frac{h}{{\lambda }_0}=\left(\frac{h}{{\lambda }^{\prime }}+\frac{h}{{\lambda }^{\prime }}\right)\cos \theta \\ =\frac{2h}{{\lambda }^{\prime }}\cos \theta \\ {\lambda }^{\prime }=\frac{2h{\lambda }_0}{h}\cos \theta \\ =2{\lambda }_0\cos \theta \end{array}$                                                                                              (X)

Write the equation for the Compton shift.

  ${\lambda }^{\prime }-{\lambda }_0=\frac{h}{m_{e}c}\left(1-\cos \theta \right)$                                                                                          (XI)

Here, $m_e$ is the mass of the electron.

Put equation (X) in equation (XI).

  $2{\lambda }_0\cos \theta -{\lambda }_0=\frac{h}{m_{e}c}\left(1-\cos \theta \right)$

Solve the above equation.

  $\begin{array}{c}2{\lambda }_0\cos \theta -{\lambda }_0=\frac{h}{m_{e}c}\left(1-\cos \theta \right)\\ =\frac{h}{m_{e}c}-\frac{h}{m_{e}c}\cos \theta \\ \left(2{\lambda }_0+\frac{h}{m_{e}c}\right)\cos \theta ={\lambda }_0+\frac{h}{m_{e}c}\end{array}$                                                                     (XII)

Rewrite equation (II) for ${\lambda }_0$ .

  ${\lambda }_0=\frac{hc}{E_0}$

Put the above equation in equation (XII).

  $\begin{array}{c}\left(2\frac{hc}{E_0}+\frac{h}{m_{e}c}\right)\cos \theta =\frac{hc}{E_0}+\frac{h}{m_{e}c}\\ \frac{2h{c}^2{m}_e+h{E}_0}{m_{e}c{E}_0}\cos \theta =\frac{h{c}^2{m}_e+h{E}_0}{m_{e}c{E}_0}\\ \frac{1}{m_{e}c{E}_0}\left(2c^2{m}_e+E_0\right)\cos \theta =\frac{1}{m_{e}c{E}_0}\left(c^2{m}_e+E_0\right)\\ \cos \theta =\frac{m_e{c}^2+E_0}{2m_e{c}^2+E_0}\end{array}$

Rewrite the above equation for $\theta$ .

  $\theta ={\cos }^{-1}\frac{m_e{c}^2+E_0}{2m_e{c}^2+E_0}$                                                                                     (XIII)

Conclusion:

It is given that the energy of the incident photon is $0.880\text{ MeV}$ .

The value of the $m_e{c}^2$ is $0.511\text{ MeV}$ .

Substitute $0.511\text{ MeV}$ for $m_e{c}^2$ and $0.880\text{ MeV}$ for $E_0$ in equation (XIII) to find $\theta$ .

  $\begin{array}{c}\theta ={\cos }^{-1}\frac{0.511\text{ MeV}+0.880\text{ MeV}}{2\left(0.511\text{ MeV}\right)+0.880\text{ MeV}}\\ =43.0°\end{array}$

Therefore, the scattering angle of the photon and the electron is $43.0°$ .

(b)

To determine

The energy and momentum of the scattered photon.

Answer to Problem 31P

The energy of the scattered photon is $0.602\text{ MeV}$ and the momentum is $3.21\times {10}^{22}\text{ kg}\cdot \text{m/s}$ .

Explanation of Solution

Put equation (X) in equation (V).

  $\begin{array}{c}{E}^{\prime }=\frac{hc}{2{\lambda }_0\cos \theta }\\ =\frac{hc}{{\lambda }_0\left(2\cos \theta \right)}\end{array}$

Put equation (II) in the above equation.

  $E^{\prime }=\frac{E_0}{2\cos \theta }$                                                                                                          (XIV)

Conclusion:

Substitute $0.880\text{ MeV}$ for $E_0$ and $43.0°$ for $\theta$ in equation (XIV) to find $E^{\prime }$ .

  $\begin{array}{c}{E}^{\prime }=\frac{0.880\text{ MeV}}{2\cos 43.0°}\\ =0.602\text{ MeV}\end{array}$

Substitute $0.602\text{ MeV}$ for $E^{\prime }$ and $3.00\times {10}^8\text{ m/s}$ for $c$ in equation (IV) to find $p^{\prime }$ .

  $\begin{array}{c}{p}^{\prime }=\frac{0.602\text{ MeV}\frac{1.60\times {10}^{-13}\text{ J}}{1\text{ MeV}}}{3.00\times {10}^8\text{ m/s}}\\ =3.21\times {10}^{22}\text{ kg}\cdot \text{m/s}\end{array}$

Therefore, the energy of the scattered photon is $0.602\text{ MeV}$ and the momentum is $3.21\times {10}^{22}\text{ kg}\cdot \text{m/s}$ .

(c)

To determine

The kinetic energy and momentum of the scattered electron.

Answer to Problem 31P

The kinetic energy of the scattered electron is $0.278\text{ MeV}$ and its momentum is $3.21\times {10}^{-22}\text{ kg}\cdot \text{m/s}$ .

Explanation of Solution

Write the equation for the kinetic energy of the scattered electron.

  $K_e=E_0-E^{\prime }$                                                                                                        (XV)

Here, $K_e$ is the kinetic energy of the scattered electron.

Conclusion:

Substitute $0.880\text{ MeV}$ for $E_0$ and $0.602\text{ MeV}$ for $E^{\prime }$ in equation (XV) to find $K_e$ .

  $\begin{array}{c}{K}_e=0.880\text{ MeV}-0.602\text{ MeV}\\ =0.278\text{ MeV}\end{array}$

Substitute $3.21\times {10}^{-22}\text{ kg}\cdot \text{m/s}$ for $p^{\prime }$ in equation (VIII) to find $p_e$ .

  $p_e=3.21\times {10}^{-22}\text{ kg}\cdot \text{m/s}$

Therefore, the kinetic energy of the scattered electron is $0.278\text{ MeV}$ and its momentum is $3.21\times {10}^{-22}\text{ kg}\cdot \text{m/s}$ .