Chapter 27, Problem 85P

To determine

The wavelengths of incident and scattered x-rays.

Answer to Problem 85P

The wavelengths of incident and scattered x-rays are $171\text{pm}$ and $176\text{pm}$.

Explanation of Solution

Write the relation for the Compton shift in wavelength of x-ray.

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

Here, ${\lambda }^{\prime }$ is wavelength of the scattered x-ray, $\lambda$ is wavelength of the incident x-ray, $h$ is Planck’s constant, $m_{\text{e}}$ is mass of the electron, $c$ is speed of the x-ray photon, and $\theta$ is scattering angle.

The term $h/m_{\text{e}}c$ in equation (1) is called Compton wavelength has the value of $2.426\text{pm}$.

Rearrange equation (1) for the wavelength of the scattered x-ray.

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

Write the relation for the kinetic energy of the electron.

  $K_{\text{e}}=E-E^{\prime }$                                                                                                              (3)

Here, $K_{\text{e}}$ is kinetic energy of the electron, $E$ is kinetic energy of incident x-ray photon, and $E^{\prime }$ is kinetic energy of scattered x-ray photon.

The change in kinetic energy of the x-ray photon is the kinetic energy of the electron.

Substitute $hc/\lambda$ for $E$ and $hc/{\lambda }^{\prime }$ for $E^{\prime }$ in equation (3) and simplify.

  $\begin{array}{l}{K}_{\text{e}}=\frac{hc}{\lambda }-\frac{hc}{{\lambda }^{\prime }}\\ \frac{K_{\text{e}}}{hc}=\frac{1}{\lambda }-\frac{1}{{\lambda }^{\prime }}\end{array}$                                                                                                           (4)

Substitute equation (2) in (4) to find ${\lambda }^{\prime }$.

  $\frac{K_{\text{e}}}{hc}=\frac{1}{\left[{\lambda }^{\prime }-\frac{h}{m_{\text{e}}c}\left(1-\cos \theta \right)\right]}-\frac{1}{{\lambda }^{\prime }}$

Substitute $2.426\text{pm}$ for $h/m_{\text{e}}c$ and $180°$ for $\theta$, $0.20\text{keV}$ for $K_{\text{e}}$, $1240\text{eV}\cdot \text{nm}$ for $hc$ to find ${\lambda }^{\prime }$.

  $\begin{array}{l}\frac{\left(0.20\text{keV}\right)\left(\frac{{10}^3\text{eV}}{1\text{keV}}\right)}{\left(1240\text{eV}\cdot \text{nm}\right)\left(\frac{{10}^3\text{pm}}{\text{nm}}\right)}=\frac{1}{\left[{\lambda }^{\prime }-\left(2.426\text{pm}\right)\left(1-\cos 180°\right)\right]}-\frac{1}{{\lambda }^{\prime }}\\ {{\lambda }^{\prime }}^2-\left(4.852\text{pm}\right){\lambda }^{\prime }=3.008\times {10}^4{\text{pm}}^{\text{2}}\\ {{\lambda }^{\prime }}^2-\left(4.852\text{pm}\right){\lambda }^{\prime }-3.008\times {10}^4{\text{pm}}^{\text{2}}=0\end{array}$                            (5)

Solve the quadratic equation in (5) to find ${\lambda }^{\prime }$.

  $\begin{array}{c}{\lambda }^{\prime }=\frac{\left(4.852\text{pm}\right)\pm \sqrt{{\left(-4.852\text{pm}\right)}^2-\left(4\right)\left(1\right)\left(-3.008\times {10}^4{\text{pm}}^{\text{2}}\right)}}{\left(2\right)\left(1\right)}\\ =176\text{pm},-171\text{pm}\end{array}$

The value of wavelength of scattered x-ray photon is $176\text{pm}$ and the negative solution is not taken into account due to the wavelength never be negative.

Rewrite equation (2) for the wavelength of the incident x-ray photon.

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

Substitute $2.426\text{pm}$ for $h/m_{\text{e}}c$ and $180°$ for $\theta$, and $176\text{pm}$ for ${\lambda }^{\prime }$ to find $\lambda$.

  $\begin{array}{c}\lambda =\left(176\text{p}\right)\text{m}-\left(2.426\text{pm}\right)\left(1-\cos 180°\right)\\ =171\text{pm}\end{array}$

Thus, the wavelength of the incident x-ray photon is $\text{171}\text{pm}$.

Conclusion:

Therefore, the wavelengths of incident and scattered x-rays are $171\text{pm}$ and $176\text{pm}$.