Chapter 26, Problem 31E

(a)

To determine

The momentum of an X-ray photon.

Answer to Problem 31E

Themomentum of an X-ray photonis $\underset{\_}{1.57\times {10}^{-27}\text{kg}}$.

Explanation of Solution

Given that the energy of an X-ray photon is $100,000\text{eV}$.

Write the expression for the momentum of photon in terms of energy.

  $p=\frac{E}{c}$        (I)

Here, $p$ is the momentum, and $E$ is the energy, and $c$ is the speed of light.

Conclusion:

Substitute $100,000\text{eV}$ for $E$ and $3\times {10}^8\text{m/s}$ for $c$ in Equation (I) to find the momentum of an X-ray photon.

  $\begin{array}{c}p=\frac{100,000\text{eV}}{3\times {10}^8\text{m/s}}\\ =3.33\times {10}^{-4}\text{eV}\cdot \text{s}\cdot {\text{m}}^{-1}\left(\frac{1.602\times {10}^{-19}\text{J}}{1\text{eV}}\right)\\ =5.34\times {10}^{-23}\text{kg}\cdot \text{m}\cdot {\text{s}}^{-1}\end{array}$

Therefore, the momentum of an X-ray photonis $\underset{\_}{5.34\times {10}^{-23}\text{kg}\cdot \text{m}\cdot {\text{s}}^{-1}}$.

(b)

To determine

The recoil energy of the electron when 5 percent of its momentum is transferred to an electron.

Answer to Problem 31E

The recoil energy of the electron when 5 percent of its momentum is transferred to an electronis $\underset{\_}{24.48\text{eV}}$.

Explanation of Solution

Write the expression for kinetic energy of the recoiling electron in terms of momentum.

  $K=\frac{p_e^2}{2m}$        (II)

Here, $K$ is the kinetic energy of the recoiling electron, $p_e$ is the momentum of electron, and $m$ is the mass of the electron.

Conclusion:

From part (a) the momentum of an X-ray photon is $5.34\times {10}^{-23}\text{kg}\cdot \text{m}\cdot {\text{s}}^{-1}$.

When 5 percent of its momentum transferred to an electron, then the momentum of the electron will be,

  $\begin{array}{c}{p}_{\text{e}}=5\%\text{ of }p\\ =\left(\frac{5}{100}\right)\left(5.34\times {10}^{-23}\text{kg}\cdot \text{m}\cdot {\text{s}}^{-1}\right)\\ =2.67\times {10}^{-24}\text{kg}\cdot \text{m}\cdot {\text{s}}^{-1}\end{array}$

Substitute $2.67\times {10}^{-24}\text{kg}\cdot \text{m}\cdot {\text{s}}^{-1}$ for $p_{\text{e}}$ and $9.1\times {10}^{-31}\text{kg}$ for $m$ in Equation (II) to find the recoil energy of the electron when 5 percent of its momentum is transferred to an electron.

  $\begin{array}{c}K=\frac{{\left(2.67\times {10}^{-24}\text{kg}\cdot \text{m}\cdot {\text{s}}^{-1}\right)}^2}{2\left(9.1\times {10}^{-31}\text{kg}\right)}\\ =3.91\times {10}^{-18}\text{J}\left(\frac{1.0\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =24.48\text{eV}\end{array}$

Therefore, the recoil energy of the electron when 5 percent of its momentum is transferred to an electronis $\underset{\_}{24.48\text{eV}}$.

(c)

To determine

The energy of the recoiling photon.

Answer to Problem 31E

The energy of the recoiling photonis $\underset{\_}{99975.52\text{eV}}$.

Explanation of Solution

From part (b) the recoil energy of the electron when 5 percent of its momentum is transferred to an electronis $24.48\text{eV}$.

Write the mathematical expression for the kinetic energy of a recoiled photon from Compton scattering.

  $E_{\text{p}}=E-K$        (III)

Here, $E_{\text{p}}$ is the energy of a recoiled photon.

Conclusion:

Substitute $100,000\text{eV}$ for $E$ and $24.48\text{eV}$ for $K$ in Equation (III) to find the energy of the recoiling photon.

  $\begin{array}{c}{E}_{\text{p}}=100,000\text{eV}-24.48\text{eV}\\ =99975.52\text{eV}\end{array}$

Therefore, the energy of the recoiling photonis $\underset{\_}{99975.52\text{eV}}$.

(d)

To determine

The decrease in frequency of the scattered electron.

Answer to Problem 31E

The decrease in frequency of the scattered electronis $\underset{\_}{5.919\times {10}^{15}\text{Hz}}$.

Explanation of Solution

From part (b) the recoil energy of the electron when 5 percent of its momentum is transferred to an electronis $24.48\text{eV}$.

Write the mathematical expression for the kinetic energy of a recoiled electron from Compton scattering is as follows:

  $K=h\left(f-f\text{'}\right)$        (IV)

Here, $f$ is the initial frequency of electron, $f\text{'}$  is the final frequency of electron, and $h$ is the Planck’s constant.

Solve the Equation (IV) for $\left(f-f\text{'}\right)$.

  $\left(f-f\text{'}\right)=\frac{K}{h}$        (V)

Conclusion:

Substitute $24.48\text{eV}$ for $K$ and $4.1357\times {10}^{-15}\text{eV}\cdot \text{s}$ for $h$ in Equation (V) to find the decrease in frequency of the scattered electron.

  $\begin{array}{c}\left(f-f\text{'}\right)=\frac{24.48\text{eV}}{4.1357\times {10}^{-15}\text{eV}\cdot \text{s}}\\ =5.919\times {10}^{15}\text{Hz}\end{array}$

Therefore, the decrease in frequency of the scattered electronis $\underset{\_}{5.919\times {10}^{15}\text{Hz}}$.