Chapter 3, Problem 48P

(a)

To determine

The wavelength if the incident photon.

Answer to Problem 48P

The wavelength if the incident photon is $0.101\text{nm}$.

Explanation of Solution

Write the formula for Compton shift,

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

Here, $m_e$ is the mass of the electron, $c$ is the speed of light and $\theta$ is the scattering angle.

Write the equation for kinetic energy of the recoiling electron using conservation of energy.

  $\frac{1}{2}{m}_e{v}^2=\frac{hc}{{\lambda }_0}-\frac{hc}{\lambda }$        (II)

Here, $m_e$ is the mass of the electron, $c$ is the speed of light and $v$ is the velocity of the photon.

Conclusion:

Substitute $9.11\times {10}^{-31}\text{ kg}$ for $m_e$, and $2.18\times {10}^6\text{m}/\text{s}$ for $v$ in expression (II)

  $\begin{array}{c}K=\frac{1}{2}{m}_e{v}^2\\ =\frac{1}{2}\times 9.11\times {10}^{-31}\text{ kg}{\left(2.18\times {10}^6\text{m}/\text{s}\right)}^2\\ =2.16\times {10}^{-18}\text{ J}\end{array}$

Write the expression for energy lost by the photon,

  $\frac{hc}{{\lambda }_0}-\frac{hc}{{\lambda }^{\prime }}=2.16\times {10}^{-18}\text{ J}$        (III)

Substitute $9.11\times {10}^{-31}\text{ kg}$ for $m_e$, $6.63\times {10}^{-34}\text{ Js }$ for $h$ and $3\times {10}^8\text{ m}$ for $c$ in expression (I)

  $\begin{array}{c}{\lambda }^{\prime }-{\lambda }_0=\frac{h}{m_{\text{e}}c}\left(1-\cos \theta \right)\\ =\frac{6.63\times {10}^{-34}\text{ Js s}}{9.11\times {10}^{-31}\text{ kg}\left(3\times {10}^8\text{ m}\right)}\left(1-\cos 17.4°\right)\\ {\lambda }^{\prime }={\lambda }_0+1.11\times {10}^{-13}\text{ m}\end{array}$        (IV)

Substituting the expression (IV) in expression (III) and solve for ${\lambda }_0$ by substituting $6.63\times {10}^{-34}\text{ Js }$ for $h$ and $3\times {10}^8\text{ m}$ for $c$ in the expression.

  $\begin{array}{c}\frac{1}{{\lambda }_0}-\frac{1}{{\lambda }_0+0.111\text{ pm}}=\frac{2.16\times {10}^{-18}\text{ J s}}{6.63\times {10}^{-34}\text{ J s}\left(\text{3}\times {\text{10}}^{\text{8}}\text{ m}\right)}=\frac{1.09\times {10}^7}{\text{m}}\\ \frac{{\lambda }_0+0.111\text{ pm}-{\lambda }_0}{{\lambda }_0^2+{\lambda }_0\left(0.111\text{ pm}\right)}=\frac{1.09\times {10}^7}{\text{m}}\\ 0.111\text{ pm}=\left(\frac{1.09\times {10}^7}{\text{m}}\right){\lambda }_0^2+1.21\times {10}^{-6}{\lambda }_0\\ 0=\left(\begin{array}{c}1.09\times {10}^7{\lambda }_0^2+1.21\times {10}^{-6}\text{ m}{\lambda }_0\\ -1.11\times {10}^{-13}{\text{ m}}^{\text{2}}\end{array}\right)\\ {\lambda }_0=\frac{-1.21\times {10}^{-6}\text{ m}\pm \left(\sqrt{\begin{array}{l}{\left(1.21\times {10}^{-6}\text{ m}\right)}^2\\ -4\left(1.09\times {10}^7\right)\left(-1.11\times {10}^{-13}{\text{ m}}^{\text{2}}\right)\end{array}}\right)}{2\left(1.09\times {10}^7\right)}\end{array}$

The wavelength if the incident photon is $0.101\text{nm}$.

(b)

To determine

The angle through which the electron scatters.

Answer to Problem 48P

The angle through which the electron scatters is $80.9°$

Explanation of Solution

Write the expression for conservation of momentum in the $y-axis$ ,

  $p_e\sin \varphi =p\sin \theta$        (I)

Write the expression for the wavelength of the scattered photon.

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

Here, $m_e$ is the mass of the electron, $c$ is the speed of light and $\theta$ is the scattering angle.

Conclusion:

Substitute $17.4°$ for $\theta$,  $9.11\times {10}^{-31}\text{ kg}$ for $m_e$, $3.00\times {10}^8$ for $c$  $6.63\times {10}^{-34}\text{ Js }$ for $h$ and $1.01\times {10}^{-10}\text{m}$ for $\lambda$ in expression (II)

  $\begin{array}{c}\lambda =\frac{6.63\times {10}^{-34}\text{ Js }}{9.11\times {10}^{-31}\text{ kg}\left(3.00\times {10}^8\right)c}\left(1-\cos 17.4°\right)+1.01\times {10}^{-10}\text{m}\\ =1.011\times {10}^{-10}\text{m}\end{array}$

The scattering angle for the electron is,

  $\varphi ={\sin }^{-1}\left(\frac{h\sin \theta }{\lambda m_{e}v}\right)$

Substitute $17.4°$ for $\theta$,  $9.11\times {10}^{-31}\text{ kg}$ for $m_e$, $3.00\times {10}^8$ for $c$  $6.63\times {10}^{-34}\text{ Js }$ for $h$, $2.18\times {10}^6\text{m/s}$ for $v$  and $1.01\times {10}^{-10}\text{m}$ for $\lambda$ in expression (II)

  $\begin{array}{c}\varphi ={\sin }^{-1}\left(\frac{\left(6.63\times {10}^{-34}\text{ Js }\right)\sin \left(17.4°\right)}{\left(1.01\times {10}^{-10}\text{m}\right)\left(9.11\times {10}^{-31}\text{ kg}\right)\left(2.18\times {10}^6\text{m/s}\right)}\right)\\ =80.9°\end{array}$

The angle through which the electron scatters is $80.9°$