Chapter 28, Problem 18P

(a)

To determine

The scattering angle of the photon and the electron.

Answer to Problem 18P

The scattering angle of the photon and the electron is $\underset{\_}{\frac{m_e{c}^2+E_0}{2m_e{c}^2+E_0}}$.

Explanation of Solution

The momentum of the photon before scattering is given by,

    $p_0=\frac{h}{{\lambda }_0}$        (I)

Here, $p_0$ is the momentum before scattering, $h$ is the Planck’s constant, ${\lambda }_0$ is the wavelength.

The momentum after scattering is given by,

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

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

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

Here, $p_e$ is the electron momentum after scattering.

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

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

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

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

Neglecting the trivial solution will gives,

    $p_e=p^{\prime }$        (VI)

Use equation (VI) in equation (IV),and solve for ${\lambda }^{\prime }$

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

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

Use equation (VII) in equation (VIII),

    $\begin{array}{c}{\lambda }^{\prime }-{\lambda }_0=\frac{h}{m_{e}c}\left(1-\cos \theta \right)\\ \left(2{\lambda }_0\cos \theta \right)-{\lambda }_0=\frac{h}{m_{e}c}\left(1-\cos \theta \right)\\ \left(2{\lambda }_0+\frac{h}{m_{e}c}\right)\cos \theta ={\lambda }_0+\frac{h}{m_{e}c}\\ \left(2\frac{h}{m_{e}c}+\frac{h}{m_{e}c}\right)\cos \theta =\frac{hc}{E_0}+\frac{hc}{m_{e}c}\end{array}$        (IX)

Solve equation (IX) for $\theta$.

  $\begin{array}{c}\frac{1}{m_e{c}^2{E}_0}\left(2m_e{c}^2+E_0\right)\cos \theta =\frac{1}{m_e{c}^2{E}_0}\left(m_e{c}^2+E_0\right)\\ \cos \theta =\frac{m_e{c}^2+E_0}{2m_e{c}^2+E_0}\end{array}$        (X)

Conclusion:

Solve equation (IX) for $\theta$.

  $\begin{array}{c}\frac{1}{m_e{c}^2{E}_0}\left(2m_e{c}^2+E_0\right)\cos \theta =\frac{1}{m_e{c}^2{E}_0}\left(m_e{c}^2+E_0\right)\\ \cos \theta =\frac{m_e{c}^2+E_0}{2m_e{c}^2+E_0}\end{array}$        (X)

Therefore, the scattering angle of the photon and the electron is $\underset{\_}{\frac{m_e{c}^2+E_0}{2m_e{c}^2+E_0}}$

(b)

To determine

The energy and the momentum of the scattered photon.

Answer to Problem 18P

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

Explanation of Solution

Write the expression for the energy of the scattered photon.

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

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

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

Use equation (VII) in equation (XII),

    $\begin{array}{c}{E}^{\prime }=\frac{hc}{{\lambda }_0\left(2\cos \theta \right)}\\ =\frac{E_0}{2\cos \theta }\end{array}$        (XIII)

Conclusion:

Use equation (X) in equation (XIII),

    $\begin{array}{c}{E}^{\prime }=\frac{E_0}{2\frac{m_e{c}^2+E_0}{2m_e{c}^2+E_0}}\\ =\frac{E_0\left(2m_e{c}^2+E_0\right)}{2\left(m_e{c}^2+E_0\right)}\end{array}$        (XIV)

Use equation (XIV) in equation (XII),

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

Therefore, The energy and the momentum of the scattered photon is $\underset{\_}{\frac{E_0\left(2m_e{c}^2+E_0\right)}{2\left(m_e{c}^2+E_0\right)}}$ and $\underset{\_}{\frac{E_0\left(2m_e{c}^2+E_0\right)}{2c\left(m_e{c}^2+E_0\right)}}$.

(c)

To determine

Kinetic energy and momentum of the scattered electron.

Answer to Problem 18P

Kinetic energy and momentum of the scattered electron is $\underset{\_}{0.278\text{MeV}}$ and $\underset{\_}{3.21\times {10}^{-22}\text{kg}\cdot \text{m}/\text{s}}$

Explanation of Solution

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

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

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

Use equation (XIV) in equation (XVI),

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

Solve the above equation,

    $\begin{array}{c}{K}_e=\frac{2E_0{m}_e{c}^2+2E_0^2-2E_0{m}_e{c}^2-E_0^2}{2\left(m_e{c}^2+E_0\right)}\\ =\frac{E_0^2}{2\left(m_e{c}^2+E_0\right)}\end{array}$        (XVII)

  $p_e=p^{\prime }$        (XVII)

Conclusion:

Use equation (XV) in equation (XVII) to find $p_e$.

    $p_e=\frac{E_0\left(2m_e{c}^2+E_0\right)}{2c\left(m_e{c}^2+E_0\right)}$

Therefore, Kinetic energy and momentum of the scattered electron is $\underset{\_}{\frac{E_0^2}{2\left(m_e{c}^2+E_0\right)}}$ and $\underset{\_}{\frac{E_0\left(2m_e{c}^2+E_0\right)}{2c\left(m_e{c}^2+E_0\right)}}$.