Chapter 27, Problem 82P

To determine

The energy of the $\gamma$-rays scattered through $\theta =90°$ and $\theta =180°$.

Answer to Problem 82P

The energy of the $\gamma$-rays scattered through $\theta =90°$ is $136\text{keV}$, and that through $\theta =180°$ is $\underset{\_}{\text{108}\text{keV}}$

Explanation of Solution

Given that the energy of the photons is $186\text{keV}$.

In Compton scattering, photons scattered from a target have longer wavelengths than the incident photons. The Compton shift is given by the expression;

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

Here, $\Delta \lambda$ is the Compton shift, $h$ is the Planck’s constant, $c$ is the speed of light, $m_{\text{e}}$ is the mass of the electron, $\theta$ is the scattering angle.

Write the expression for the wavelength of an incident photon.

    $\lambda =\frac{hc}{E}$                                                                                                                    (II)

Here, $E$ is the energy of the photon.

Write the expression for the wavelength of the scattered x-rays.

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

Here, ${\lambda }^{\prime }$ is the wavelength of the photon scattered at angle $\theta$, $\lambda$ is the wavelength of the incident photon.

Write the expression for the energy of the scattered ray.

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

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

Conclusion:

Substitute $90°$ for $\theta$, and $2.426\text{pm}$ for $\frac{h}{m_{e}c}$ in equation (I) to find $\Delta {\lambda }_{90°}$.

  $\begin{array}{c}\Delta {\lambda }_{90°}=2.426\text{pm}\left(1-\cos 90°\right)\\ =2.426\text{pm}\end{array}$

Substitute $180°$ for $\theta$, and $2.426\text{pm}$ for $\frac{h}{m_{e}c}$ in equation (I) to find $\Delta {\lambda }_{180°}$.

  $\begin{array}{c}\Delta {\lambda }_{180°}=2.426\text{pm}\left(1-\cos 180°\right)\\ =4.852\text{pm}\end{array}$

Substitute $1240\text{eV}\cdot \text{nm}$ for $hc$, $186\text{keV}$ for $E$ in equation (II) to find $\lambda$.

  $\begin{array}{c}\lambda =\frac{1240\text{eV}\cdot \text{nm}}{186\text{keV}}\\ =\frac{\left(1240\text{eV}\cdot \text{nm}\times \frac{1\text{m}}{1\times {10}^9\text{nm}}\right)}{\left(186\text{keV}\times \frac{1000\text{eV}}{1\text{keV}}\right)}\\ =6.667\times {10}^{-12}\text{m}\times \frac{1\text{pm}}{1\times {10}^{-12}\text{m}}\\ =6.667\text{pm}\end{array}$

Substitute $6.667\text{pm}$ for $\lambda$, $2.426\text{pm}$ for $\Delta \lambda$ in equation (III) to0 find ${\lambda }_{90}{}^{\prime }$.

    $\begin{array}{c}{\lambda }_{90}{}^{\prime }=6.667\text{pm+2}\text{.426}\text{pm}\\ =9.09\text{pm}\end{array}$

Substitute $6.667\text{pm}$ for $\lambda$, $4.852\text{pm}$ for $\Delta \lambda$ in equation (III) and solve for ${\lambda }_{180°}{}^{\prime }$.

    $\begin{array}{c}{\lambda }_{180°}{}^{\prime }=6.667\text{pm+4}\text{.852}\text{pm}\\ \text{=11}\text{.52}\text{pm}\end{array}$

Substitute $1240\text{eV}\cdot \text{nm}$ for $hc$ and $9.09\text{pm}$ for ${{\lambda }^{\prime }}_{90°}$ in equation (IV) to find $E_{90°}$.

    $\begin{array}{c}{E}_{90°}=\frac{1240\text{eV}\cdot \text{nm}}{9.09\text{pm}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{9.09\text{pm}\times \frac{1\text{nm}}{1000\text{pm}}}\\ =1.36\times {10}^5\text{eV}\times \frac{1\text{keV}}{1000\text{eV}}\\ =136\text{keV}\end{array}$

Substitute $1240\text{eV}\cdot \text{nm}$ for $hc$ and $11.52\text{pm}$ for ${{\lambda }^{\prime }}_{180°}$ in equation (IV) to find $E_{180°}$.

    $\begin{array}{c}{E}_{180°}=\frac{1240\text{eV}\cdot \text{nm}}{11.52\text{pm}}\\ =\frac{1240\text{eV}\cdot \text{nm}}{11.52\text{pm}\times \frac{1\text{nm}}{1000\text{pm}}}\\ =1.08\times {10}^5\text{eV}\times \frac{1\text{keV}}{1000\text{eV}}\\ =108\text{keV}\end{array}$

Therefore, the energy of the $\gamma$-rays scattered through $\theta =90°$ is $136\text{keV}$, and that through $\theta =180°$ is $\underset{\_}{\text{108}\text{keV}}$.