Chapter 39, Problem 19P

(a)

To determine

The scattering angle of the proton and electron.

Answer to Problem 19P

The scattering angle of the proton and electron is $43.0°$.

Explanation of Solution

Write the expression for the conservation of momentum along horizontal direction.

  $\frac{h}{\lambda }=\frac{h}{{\lambda }^{\prime }}\cos \theta +p\cos \theta$                                                                                            (I)

Here, $\lambda$ is wavelength of the photon before collision, ${\lambda }^{\prime }$ is the wavelength of the photon after collision, p is the momentum of electron and $\theta$ is the direction scattering for both electron and photon.

Write the expression for the conservation of momentum along the vertical direction.

  $0=\frac{h}{{\lambda }^{\prime }}\sin \theta -p\sin \theta$

Rewrite the above equation in terms of p.

  $\begin{array}{c}\frac{h}{{\lambda }^{\prime }}\sin \theta =p\sin \theta \\ p=\frac{h}{{\lambda }^{\prime }}\end{array}$

Substitute the above expression in (I) to rewrite.

  $\begin{array}{c}\frac{h}{\lambda }=\frac{h}{{\lambda }^{\prime }}\cos \theta +\left(\frac{h}{{\lambda }^{\prime }}\right)\cos \theta \\ \frac{1}{\lambda }=\frac{2}{{\lambda }^{\prime }}\cos \theta \\ \cos \theta =\frac{{\lambda }^{\prime }}{2\lambda }\end{array}$                                                                                    (I)

Write the expression for the Compton scattering.

  ${\lambda }^{\prime }=\lambda +\frac{h}{mc}\left(1-\cos \theta \right)$

Here, m is the mass of the electron.

Substitute the above expression in (II) to rewrite.

  $\begin{array}{c}\cos \theta =\frac{\lambda +\frac{h}{mc}\left(1-\cos \theta \right)}{2\lambda }\\ 2\lambda \cos \theta =\lambda +\frac{h}{mc}-\frac{h}{mc}\cos \theta \\ \left(2\lambda +\frac{h}{mc}\right)\cos \theta =\lambda +\frac{h}{mc}\end{array}$                                                                    (III)

Write the expression for the wavelength of the photon before collision.

  $\lambda =\frac{hc}{E_0}$

Here, $E_0$ is the energy of the photon.

Substitute the above equation in (III) to rewrite in terms of $\theta$.

  $\begin{array}{c}\left(2\left(\frac{hc}{E_0}\right)+\frac{h}{mc}\right)\cos \theta =\frac{hc}{E_0}+\frac{h}{mc}\\ \frac{1}{E_{0}mc}\left(2m{c}^2+E_0\right)\cos \theta =\frac{1}{E_{0}mc}\left(m{c}^2+E_0\right)\\ \cos \theta =\frac{m{c}^2+E_0}{2m{c}^2+E_0}\\ \theta ={\cos }^{-1}\left(\frac{m{c}^2+E_0}{2m{c}^2+E_0}\right)\end{array}$

Write the expression to calculate the rest mass energy of electron.

  $E=m{c}^2$

Here, E is the rest mass energy of the photon.

Substitute the above equation in the expression for $\theta$ to rewrite.

  $\theta ={\cos }^{-1}\left(\frac{E+E_0}{2E+E_0}\right)$

The rest mass energy of the photon is $0.511\text{MeV}$.

Substitute $0.511\text{MeV}$ for E and $0.880\text{MeV}$ for $E_0$ in the above equation to calculate $\theta$.

  $\begin{array}{c}\theta ={\cos }^{-1}\left(\frac{0.880\text{MeV}+0.511\text{MeV}}{2\left(0.511\text{MeV}\right)+0.880\text{MeV}}\right)\\ ={\cos }^{-1}\left(\frac{1.391}{1.902}\right)\\ =43.0°\end{array}$

Conclusion:

Therefore, the scattering angle of the proton and electron is $43.0°$.

(b)

To determine

The energy and momentum of the scattered photon.

Answer to Problem 19P

The energy and momentum of the scattered photon is respectively $0.600\text{MeV}$ and $3.20\times {10}^{-22}\text{kg}\cdot \text{m}/\text{s}$.

Explanation of Solution

Write the expression to calculate the energy of the scattered photon.

    $E^{\prime }=\frac{hc}{{\lambda }^{\prime }}$

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

Substitute the expression for ${\lambda }^{\prime }$ using the expression given in part (a).

    $E^{\prime }=\frac{hc}{\lambda +\frac{h}{mc}\left(1-\cos \theta \right)}$

Rewrite the above expression using $\lambda =\frac{hc}{E_0}$ in terms of $E^{\prime }$.

    $\begin{array}{l}{E}^{\prime }=\frac{hc}{\frac{hc}{E_0}+\frac{h}{mc}\left(1-\cos \theta \right)}\\ E^{\prime }=\frac{1}{\frac{1}{E_0}+\frac{h}{m{c}^2}\left(1-\cos \theta \right)}\end{array}$

Rewrite the equation using $E=m{c}^2$ to calculate $E^{\prime }$.

    $E^{\prime }=\frac{1}{\frac{1}{E_0}+\frac{1}{E}\left(1-\cos \theta \right)}$

Substitute $0.511\text{MeV}$ for E, $43.0°$ for $\theta$ and $0.880\text{MeV}$ for $E_0$ in the above equation to calculate $E^{\prime }$.

    $\begin{array}{c}{E}^{\prime }=\frac{1}{\frac{1}{0.880\text{MeV}}+\frac{1}{0.511\text{MeV}}\left(1-\cos 43.0°\right)}\\ =\frac{1}{\left(1.14+0.526\right){\text{MeV}}^{-1}}\\ =0.600\text{MeV}\end{array}$

Write the expression to calculate the momentum of the scattered electron.

    $p^{\prime }=\frac{E^{\prime }}{c}$

Here, $p^{\prime }$ is the momentum of the scattered electron and c is the speed of light.

Substitute $0.600\text{MeV}$ for $E^{\prime }$ and $3.00\times {10}^8\text{m}/\text{s}$ for c in the above equation to calculate $p^{\prime }$.

    $\begin{array}{c}{p}^{\prime }=\frac{0.600\text{MeV}\left(\frac{{10}^6\text{eV}}{1\text{MeV}}\right)\left(\frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}{3.00\times {10}^8\text{m}/\text{s}}\\ =3.20\times {10}^{-22}\text{kg}\cdot \text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the energy and momentum of the scattered photon is respectively $0.600\text{MeV}$ and $3.20\times {10}^{-22}\text{kg}\cdot \text{m}/\text{s}$.

(c)

To determine

The kinetic energy and momentum of the scattered electron.

Answer to Problem 19P

The energy and momentum of the scattered electron is respectively $\text{0}\text{.280}\text{MeV}$ and $3.20\times {10}^{-22}\text{kg}\cdot \text{m}/\text{s}$.

Explanation of Solution

Write the expression to calculate the kinetic energy of the scattered electron.

    $K=E_0-E^{\prime }$

Here, K is the kinetic energy of the scattered electron.

Substitute $0.600\text{MeV}$ for $E^{\prime }$ and $0.880\text{MeV}$ for $E_0$ in the above equation to calculate K.

    $\begin{array}{c}K=0.880\text{MeV}-0.600\text{MeV}\\ =\text{0}\text{.280}\text{MeV}\end{array}$

Refer the equation for p in part (a) to calculate the momentum of the electron.

Write the expression for the momentum of the electron.

    $p=\frac{h}{{\lambda }^{\prime }}$

Substitute the expression for ${\lambda }^{\prime }$ using the expression given in part (a) to rewrite the above equation.

    $p=\frac{h}{\lambda +\frac{h}{mc}\left(1-\cos \theta \right)}$

Rewrite the above expression using $\lambda =\frac{hc}{E_0}$ in terms of p.

    $\begin{array}{c}p=\frac{h}{\frac{hc}{E_0}+\frac{h}{mc}\left(1-\cos \theta \right)}\\ p=\frac{1}{\frac{c}{E_0}+\frac{1}{mc}\left(1-\cos \theta \right)}\end{array}$

Substitute $0.880\text{MeV}$ for $E_0$, $3.00\times {10}^8\text{m}/\text{s}$ for c, $43.0°$ for $\theta$ and $9.11\times {10}^{-31}\text{kg}$ for m in the above equation to calculate p.

    $\begin{array}{c}p=\frac{1}{\frac{3.00\times {10}^8\text{m}/\text{s}}{0.880\text{MeV}\left(\frac{{10}^6\text{eV}}{1\text{MeV}}\right)\left(\frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)}+\frac{1}{9.11\times {10}^{-31}\text{kg}\left(3.00\times {10}^8\text{m}/\text{s}\right)}\left(1-\cos 43.0°\right)}\\ =\frac{1}{0.0213\times {10}^{-23}{\left(\text{kg}\cdot \text{m}/\text{s}\right)}^{-1}+0.00984\times {10}^{-23}{\left(\text{kg}\cdot \text{m}/\text{s}\right)}^{-1}}\\ =3.21\times {10}^{-22}\text{kg}\cdot \text{m}/\text{s}\\ \approx 3.20\times {10}^{-22}\text{kg}\cdot \text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the energy and momentum of the scattered electron is respectively $\text{0}\text{.280}\text{MeV}$ and $3.20\times {10}^{-22}\text{kg}\cdot \text{m}/\text{s}$.