Chapter 3, Problem 31P

(a)

To determine

The energy of the scattered photon,

Answer to Problem 31P

The energy of the scattered photon is $9.11\times {10}^4\text{ eV}$

Explanation of Solution

Write the equation for the final energy,

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

Here, ${\lambda }^{\prime }$ is the final wavelength, $h$ is the Planck’s constant, $E^{\prime }$ the final energy and $c$ is the speed of light.

Write the expression for wavelength in terms of initial wavelength and change in wavelength

  ${\lambda }^{\prime }={\lambda }_0+\Delta \lambda$        (II)

Here, ${\lambda }^{\prime }$ is the final wavelength ${\lambda }_0$ is the initial wavelength and $\Delta \lambda$ is the change in wavelength.

Write the equation for the initial energy,

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

Here, ${\lambda }_0$ is the wavelength, $h$ is the Planck’s constant, $E_0$ the initial energy and $c$ is the speed of light.

Write the equation for the difference in wavelength,

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

Here, $\Delta \lambda$ is the difference in wavelength, $h$ is the Planck’s constant, $m_e$ the mass of the electron and $c$ is the speed of light.

Conclusion:

Substitute $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $h$ , $3\times {10}^8\text{m}/\text{s}$ for $c$ and $0.1\text{ MeV}$ for $E_0$ in equation (III).

    $\begin{array}{c}{\lambda }_0=\frac{hc}{E_0}\\ =\frac{\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{0.1\text{ MeV}}\\ =1.243\times {10}^{-11}\text{ m}\end{array}$

Substitute $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $h$ $9.11\times {10}^{-34}\text{ kg}$ for $m_e$, $3\times {10}^8\text{m}/\text{s}$ for $c$ and $60°$ for $\theta$ in equation (IV).

  $\begin{array}{c}\Delta \lambda =\left(\frac{h}{m_{e}c}\right)\left(1-\cos \theta \right)\\ =\frac{\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(1-\cos 60°\right)}{\left(9.11\times {10}^{-34}\text{ kg}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}\\ =1.215\times {10}^{-12}\text{ m}\end{array}$

Substitute $1.243\times {10}^{-11}\text{ m}$ for ${\lambda }_0$ and $1.215\times {10}^{-12}\text{ m}$ for $\Delta \lambda$ in expression (II)

  $\begin{array}{c}{\lambda }^{\prime }={\lambda }_0+\Delta \lambda \\ =1.364\times {10}^{-11}\text{ m}\end{array}$

Substitute $6.63\times {10}^{-34}\text{ J}\cdot \text{s}$ for $h$ $1.364\times {10}^{-11}\text{ m}$ for ${\lambda }^{\prime }$, and $3\times {10}^8\text{m}/\text{s}$ for $c$ and in equation (I).

  $\begin{array}{c}{E}^{\prime }=\frac{hc}{{\lambda }^{\prime }}\\ =\frac{\left(6.63\times {10}^{-34}\text{ J}\cdot \text{s}\right)\left(3\times {10}^8\text{m}/\text{s}\right)}{1.364\times {10}^{-11}\text{ m}}\\ =9.11\times {10}^4\text{ eV}\end{array}$

Thus, the energy of the scattered photon is $9.11\times {10}^4\text{ eV}$

(b)

To determine

The recoil kinetic energy of the electron.

Answer to Problem 31P

The recoil kinetic energy of the electron is $8.90\text{ keV}$.

Explanation of Solution

Write the equation for the final energy,

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

Here, ${\lambda }^{\prime }$ is the final wavelength, $h$ is the Planck’s constant, $E^{\prime }$ the final energy and $c$ is the speed of light.

Write the equation for the initial energy,

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

Here, ${\lambda }_0$ is the wavelength, $h$ is the Planck’s constant, $E_0$ the initial energy and $c$ is the speed of light.

Write the expression of total energy

  $\frac{hc}{{\lambda }_0}=\frac{hc}{{\lambda }^{\prime }}+K_e$        (III)

Here, ${\lambda }_0$ is the wavelength, ${\lambda }^{\prime }$ is the final wavelength, $h$ is the Planck’s constant, $E_0$ the initial energy and $c$ is the speed of light.

Conclusion:

Substitute $0.1\text{ MeV}$ for $\frac{hc}{{\lambda }_0}$ and $91.1\text{ keV}$ for $\frac{hc}{{\lambda }^{\prime }}$ in expression (III)

    $\begin{array}{c}{K}_{\text{e}}=0.1\text{ MeV}-91.1\text{ keV}\\ =8.90\text{ keV}\end{array}$   

Therefore, the recoil kinetic energy of the electron is $8.90\text{ keV}$.

(c)

To determine

The recoil angle of the electron.

Answer to Problem 31P

The recoil angle of the electron is $55.4°$.

Explanation of Solution

Write the equation for the Conservation of momentum along $x$ ,

    $\frac{h}{{\lambda }_0}=\left(\frac{h}{{\lambda }^{\prime }}\right)\cos \theta +\gamma m_{e}v\cos \varphi$        (I)

Here, ${\lambda }^{\prime }$ is the final wavelength, $h$ is the Planck’s constant, $m_e$ the mass of the electron and $v$ is the velocity.

Write the equation for the Conservation of momentum along $y$ ,

    $\left(\frac{h}{{\lambda }^{\prime }}\right)\sin \theta =\gamma m_{e}v\sin \varphi$        (II)

Here, ${\lambda }^{\prime }$ is the final wavelength, $h$ is the Planck’s constant, $m_e$ the mass of the electron, and $v$ is the velocity.

Divide the expression (I) and (II)

  $\begin{array}{c}\frac{\gamma m_{e}v\sin \varphi }{\gamma m_{e}v\cos \varphi }=\left(\frac{h}{{\lambda }^{\prime }}\right)\sin \theta \left[\left(\frac{h}{{\lambda }_0}\right)-\left(\frac{h}{{\lambda }^{\prime }}\right)\cos \theta \right]\\ \tan \varphi =\frac{{\lambda }_0\sin \theta }{\left({\lambda }^{\prime }-{\lambda }_0\right)\cos \theta }\end{array}$        (III)

Conclusion:

Substitute $1.24\times {10}^{-11}\text{ m}$ for ${\lambda }_0$, $60°$ for $\theta$, $1.36$ for ${\lambda }^{\prime }$,and $1.24$ for ${\lambda }_0$  in expression (III)

    $\begin{array}{c}\tan \varphi =\frac{\left(1.24\times {10}^{-11}\text{ m}\right)\left(\sin 60°\right)}{\left(1.36-1.24\cos 60°\right)\times {10}^{-11}\text{ m}}=1.451\\ \varphi =55.4°\end{array}$

Therefore, the recoil angle of the electron is $55.4°$.