Chapter 40, Problem 36P

(a)

To determine

The maximum fractional energy loss for the $0.511\text{ MeV}$ gamma ray Compton scattered from a free electron.

Answer to Problem 36P

The maximum fractional energy loss for the $0.511\text{ MeV}$ gamma ray Compton scattered from a free electron is $0.667$ .

Explanation of Solution

Write the Compton shift equation.

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

Here, $\Delta \lambda$ is the Compton shift in the gamma ray’s wavelength, $h$ is the Plank’s constant, $m$ is the mass of the target particle, $c$ is the speed of light in vacuum and $\theta$ is the scattering angle.

The energy of the gamma ray and its wavelength are inversely proportional. The maximum energy loss will occur for maximum increase in wavelength of the gamma ray.

The value of $\Delta \lambda$ will be maximum when the value of $\theta$ is $180°$ .

Substitute $180°$ for $\theta$ in equation (I) to find the maximum increase in wavelength.

  $\begin{array}{c}\Delta \lambda =\frac{h}{mc}\left(1-\cos 180°\right)\\ =\frac{h}{mc}\left(1-\left(-1\right)\right)\\ =\frac{2h}{mc}\end{array}$                                                                                             (II)

Write the equation for the fractional energy loss of the Gamma ray.

  $E_F=\frac{E_0-E^{\prime }}{E_0}$                                                                                                          (III)

Here, $E_F$ is the fractional energy loss, $E_0$ is the energy of the gamma ray and $E^{\prime }$ is the energy of the scattered gamma ray.

Write the equation for the energy of incident gamma ray.

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

Here, ${\lambda }_0$ is the wavelength of the incident gamma ray.

Write the equation for the energy of the scattered gamma ray.

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

Here, ${\lambda }^{\prime }$ is the wavelength of the scattered gamma ray.

Put equations (IV) and (V) in equation (III).

  $\begin{array}{c}{E}_F=\frac{hc/{\lambda }_0-hc/{\lambda }^{\prime }}{hc/{\lambda }_0}\\ =\frac{1/{\lambda }_0-1/{\lambda }^{\prime }}{1/{\lambda }_0}\\ =\frac{\left({\lambda }^{\prime }-{\lambda }_0\right){\lambda }_0}{{\lambda }_0{\lambda }^{\prime }}\\ =\frac{{\lambda }^{\prime }-{\lambda }_0}{{\lambda }^{\prime }}\end{array}$                                                                                                (VI)

Write the equation for the Compton shift in the gamma ray’s wavelength.

  $\Delta \lambda ={\lambda }^{\prime }-{\lambda }_0$                                                                                                           (VII)

Rewrite equation (VII) for ${\lambda }^{\prime }$ .

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

Put equation (VII) in the numerator of equation (VI) and replace denominator of equation (VI) by (VIII).

  $E_F=\frac{\Delta \lambda }{{\lambda }_0+\Delta \lambda }$                                                                                                          (IX)

Rewrite equation (IV) for ${\lambda }_0$.

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

Put equations (II) and (X) in equation (IX).

  $\begin{array}{c}{E}_F=\frac{2h/mc}{hc/E_0+\left(2h/mc\right)}\\ =\frac{\frac{2h}{mc}}{\frac{h\left(m{c}^2+2E_0\right)}{E_{0}mc}}\\ =\frac{2h}{mc}\times \frac{E_{0}mc}{h\left(m{c}^2+2E_0\right)}\\ =\frac{2E_0}{m{c}^2+2E_0}\end{array}$                                                                                       (XI)

Conclusion:

The value of $c$ is $3.00\times {10}^8\text{ m/s}$ and the mass of the electron is $9.11\times {10}^{-31}\text{ kg}$ .

Substitute $0.511\text{ MeV}$ for $E_0$ , $9.11\times {10}^{-31}\text{ kg}$ for $m$ and $3.00\times {10}^8\text{ m/s}$ for $c$ in equation (XI) to find fractional energy loss when the gamma ray is scattered from free electron.

  $\begin{array}{c}{E}_F=\frac{2\left(0.511\text{ MeV}\frac{{10}^6\times 1.60\times {10}^{-19}\text{ J}}{1\text{ MeV}}\right)}{\left(9.11\times {10}^{-31}\text{ kg}\right){\left(3.00\times {10}^8\text{ m/s}\right)}^2+2\left(0.511\text{ MeV}\frac{{10}^6\times 1.60\times {10}^{-19}\text{ J}}{1\text{ MeV}}\right)}\\ =0.667\end{array}$

Therefore, the maximum fractional energy loss for the $0.511\text{ MeV}$ gamma ray Compton scattered from a free electron is $0.667$ .

(b)

To determine

The maximum fractional energy loss for the $0.511\text{ MeV}$ gamma ray Compton scattered from a free proton.

Answer to Problem 36P

The maximum fractional energy loss for the $0.511\text{ MeV}$ gamma ray Compton scattered from a free proton is $0.00109$ .

Explanation of Solution

Equation (XI) can be used to find the fractional loss of the gamma ray when it is Compton scattered from the free proton by using the mass of the proton in the equation.

Conclusion:

The mass of proton is $1.6726\times {10}^{-27}\text{ kg}$

Substitute $0.511\text{ MeV}$ for $E_0$ , $1.6726\times {10}^{-27}\text{ kg}$ for $m$ and $3.00\times {10}^8\text{ m/s}$ for $c$ in equation (XI) to find fractional energy loss when the gamma ray is scattered from free proton.

  $\begin{array}{c}{E}_F=\frac{2\left(0.511\text{ MeV}\frac{{10}^6\times 1.60\times {10}^{-19}\text{ J}}{1\text{ MeV}}\right)}{\left(1.6726\times {10}^{-27}\text{ kg}\right){\left(3.00\times {10}^8\text{ m/s}\right)}^2+2\left(0.511\text{ MeV}\frac{{10}^6\times 1.60\times {10}^{-19}\text{ J}}{1\text{ MeV}}\right)}\\ =0.00109\end{array}$

Therefore, the maximum fractional energy loss for the $0.511\text{ MeV}$ gamma ray Compton scattered from a free proton is $0.00109$ .