Chapter 26, Problem 32E

(a)

To determine

The change in wavelength when a photon undergoes Compton scattering.

Answer to Problem 32E

Thechange in wavelength is $\underset{\_}{2.42\times {10}^{-12}\text{m}}$.

Explanation of Solution

Given that the wavelength of X-rays is $0.5\text{nm}$, and the scattered angle is $90°$.

The expression for the change in wavelength during Compton scattering is as follows:

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

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

Conclusion:

Substitute $6.62\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $9.1\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$, and $90°$ for $\theta$ in Equation (I) to find the change in wavelength.

  $\begin{array}{c}\Delta \lambda =\left(\frac{6.62\times {10}^{-34}\text{J}\cdot \text{s}}{\left(9.1\times {10}^{-31}\text{kg}\right)\left(3\times {10}^8\text{m/s}\right)}\right)\left(1-\cos 90°\right)\\ =2.42\times {10}^{-12}\text{m}\end{array}$

Therefore, the change in wavelength is $\underset{\_}{2.42\times {10}^{-12}\text{m}}$.

(b)

To determine

The change in frequency when a photon undergoes Compton scattering.

Answer to Problem 32E

The change in frequency is $\underset{\_}{\text{1}\text{.24}\times {10}^{20}\text{Hz}}$.

Explanation of Solution

From part (a) the change in wavelength is $2.42\times {10}^{-12}\text{m}$.

Write the expression for change in frequency.

  $\Delta f=\frac{c}{\Delta \lambda }$        (II)

Here, $\Delta f$ is the change in frequency.

Conclusion:

Substitute $2.42\times {10}^{-12}\text{m}$ for $\Delta \lambda$ and $3\times {10}^8\text{m/s}$ for $c$ in Equation (II) to find the change in frequency.

  $\begin{array}{c}\Delta f=\frac{3\times {10}^8\text{m/s}}{2.42\times {10}^{-12}\text{m}}\\ =1.239\times {10}^{20}\text{Hz}\\ \approx \text{1}\text{.24}\times {10}^{20}\text{Hz}\end{array}$

Therefore, the change in frequency is $\underset{\_}{\text{1}\text{.24}\times {10}^{20}\text{Hz}}$.

(c)

To determine

The energy acquired by the recoiling electron.

Answer to Problem 32E

The energy acquired by the recoiling electron is $\underset{\_}{1\text{1}\text{.9}\text{eV}}$.

Explanation of Solution

Write the mathematical expression for the kinetic energy of a recoiled electron from Compton scattering in terms of wavelength.

  $E_{\text{el}}=hc\left(\frac{1}{\lambda }-\frac{1}{\lambda \text{'}}\right)$        (III)

Here, $E_{\text{el}}$ is the energy of recoiling electron, $f$ is the initial frequency of electron, and $f\text{'}$  is the final frequency of electron.

Conclusion:

The wavelength of the scattered photon is equal to the sum of the wavelength of a beam of X-rays and the change in wavelength.

From part (a) the change in wavelength is $2.42\times {10}^{-12}\text{m}$.

The wavelength of the scattered photon is,

  $\lambda \text{'}=\lambda +\Delta \lambda$        (IV)

Substitute $2.42\times {10}^{-12}\text{m}$ for $\Delta \lambda$ and $0.5\text{nm}$ for $\lambda$ in Equation (IV) to find the wavelength of the scattered photon.

  $\begin{array}{c}\lambda \text{'}=\left(0.5\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)+\left(2.42\times {10}^{-12}\text{m}\right)\\ =5.0242\times {10}^{-10}\text{m}\end{array}$

Substitute $6.62\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $3\times {10}^8\text{m/s}$ for $c$, $5.0242\times {10}^{-10}\text{m}$ for  $\lambda \text{'}$, and $0.5\text{nm}$ for $\lambda$ in Equation (III) to find the energy acquired by the recoiling electron.

  $\begin{array}{c}{E}_{\text{el}}=\left(6.62\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3\times {10}^8\text{m/s}\right)\left(\frac{1}{0.5\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}}-\frac{1}{5.0242\times {10}^{-10}\text{m}}\right)\\ =1.91\times {10}^{-18}\text{J}\times \frac{1.0\text{eV}}{1.6\times {10}^{-19}\text{eV}}\\ =1\text{1}\text{.9}\text{eV}\end{array}$

Therefore, the energy acquired by the recoiling electron is $\underset{\_}{1\text{1}\text{.9}\text{eV}}$.