Chapter 27, Problem 27P

(a)

To determine

The momentum of the electron.

Answer to Problem 27P

The momentum of the electron is $1.18\times {10}^{-23}\text{kg}\text{.m/s}$.

Explanation of Solution

The figure shows the situation before and after the scattering process.

The Compton shift formula is,

$\lambda \text{'}-\lambda =\frac{h}{m_{e}c}\left(1-\cos 180°\right)=\frac{2h}{m_{e}c}$

• • ${\lambda }_0$ is the wavelength of the x-ray
  • $h$ is the Plank’s constant
  • $c$ is the speed of the light
  • $m_e$ is the mass of the electron

Substitute $0.110\text{nm}$ for $\lambda$, $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$ and $3.00\times {10}^8\text{m/s}$ for $c$.

$\begin{array}{c}\lambda \text{'}=\left(0.110\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)+\frac{2\left(6.63\times {10}^{-34}\text{J}\text{.s}\right)}{\left(9.11\times {10}^{-31}\text{kg}\right)\left(3.00\times {10}^8\text{m/s}\right)}\\ =0.115\text{nm}\end{array}$

The momentum of the recoiling electron is,

$p_e=p+p\text{'}$

• • $p$ is the momentum of the incident photon
  • $p\text{'}$ is the momentum of the scattered photon

$p_e=\frac{h}{\lambda }+\frac{h}{\lambda \text{'}}$

Substitute $0.110\text{nm}$ for $\lambda$, $0.115\text{nm}$ for $\lambda \text{'}$, $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$ and $3.00\times {10}^8\text{m/s}$ for $c$.

$\begin{array}{c}{p}_e=\left(6.63\times {10}^{-34}\text{J}\text{.s}\right)\left(\frac{1}{\left(0.110\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}+\frac{1}{\left(0.115\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\right)\\ =1.18\times {10}^{-23}\text{kg}\text{.m/s}\end{array}$

Conclusion:

Thus, the momentum of the electron is $1.18\times {10}^{-23}\text{kg}\text{.m/s}$.

(b)

To determine

The kinetic energy of the electron.

Answer to Problem 27P

The kinetic energy of the electron is $478\text{eV}$.

Explanation of Solution

The kinetic energy of the electron is,

$KE=\frac{p_e^2}{2m_e}$

Substitute $1.18\times {10}^{-23}\text{kg}\text{.m/s}$ for $p_e$ and $9.11\times {10}^{-31}\text{kg}$ for $m_e$.

$\begin{array}{c}KE=\frac{{\left(1.18\times {10}^{-23}\text{kg}\text{.m/s}\right)}^2}{2\left(9.11\times {10}^{-31}\text{kg}\right)}\left(\frac{1\text{eV}}{1.60\times {10}^{-19}\text{J}}\right)\\ =478\text{eV}\end{array}$

Conclusion:

Thus, the kinetic energy of the electron is $478\text{eV}$