Chapter 40, Problem 28P

(a)

To determine

The wavelengths of photons scattered at angles $30.0°$, $60.0°$, $90.0°$ , $120.0°$, $150.0°$, $180.0°$.

Answer to Problem 28P

The wavelengths of photons scattered at angles $30.0°$, $60.0°$, $90.0°$ , $120.0°$, $150.0°$, $180.0°$ are $120.3\times {10}^{-12}\text{m}$, $\text{121}\text{.2}\times {10}^{-12}\text{m}$, $\text{122}\text{.4}\times {10}^{-12}\text{m}$, $\text{123}\text{.6}\times {10}^{-12}\text{m}$, $\text{124}\text{.5}\times {10}^{-12}\text{m}$ and $\text{124}\text{.8}\times {10}^{-12}\text{m}$ respectively.

Explanation of Solution

Write the expression to find the change in wavelength.

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

Here, $h$ is the Plank’s constant, $m_e$ is the mass of electron, $c$ is the speed of light, $\Delta \lambda$ is the change in wavelength, $\theta$ is the scattering angle.

Write the expression to find the wavelengths of scattered photons.

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

Here, $\lambda$ is the wavelengths of scattered photons, ${\lambda }^{\prime }$ is the wavelength of X-rays before scattering.

Substitute equation (I) in (II) to find the wavelength of scattered photons.

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

Conclusion:

Substitute $120.0\text{pm}$ for ${\lambda }^{\prime }$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$, $2.998\times {10}^8\text{m/s}$ for $c$ and $30.0°$ for $\theta$ to find the wavelength of scattered photon at $30.0°$.

  $\begin{array}{c}\lambda =120.0\text{pm}\left(\frac{{10}^{-12}\text{m}}{1.0\text{pm}}\right)+\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{\left(9.11\times {10}^{-31}\text{kg}\right)\left(2.998\times {10}^8\text{m/s}\right)}\left(1-\cos \left(30.0°\right)\right)\to \\ =120.0\times {10}^{-12}\text{m}+0.3\times {10}^{-12}\text{m}\\ =120.3\times {10}^{-12}\text{m}\end{array}$

Substitute $120.0\text{pm}$ for ${\lambda }^{\prime }$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$, $2.998\times {10}^8\text{m/s}$ for $c$ and $60.0°$ for $\theta$ to find the wavelength of scattered photon at $60.0°$.

  $\begin{array}{c}\lambda =120.0\text{pm}\left(\frac{{10}^{-12}\text{m}}{1.0\text{pm}}\right)+\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{\left(9.11\times {10}^{-31}\text{kg}\right)\left(2.998\times {10}^8\text{m/s}\right)}\left(1-\cos \left(60.0°\right)\right)\\ =120.0\times {10}^{-12}\text{m}+1.2\times {10}^{-12}\text{m}\\ \text{=121}\text{.2}\times {10}^{-12}\text{m}\end{array}$

Substitute $120.0\text{pm}$ for ${\lambda }^{\prime }$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$, $2.998\times {10}^8\text{m/s}$ for $c$ and $90.0°$ for $\theta$ to find the wavelength of scattered photon at $90.0°$.

  $\begin{array}{c}\lambda =120.0\text{pm}\left(\frac{{10}^{-12}\text{m}}{1.0\text{pm}}\right)+\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{\left(9.11\times {10}^{-31}\text{kg}\right)\left(2.998\times {10}^8\text{m/s}\right)}\left(1-\cos \left(90.0°\right)\right)\\ =120.0\times {10}^{-12}\text{m}+2.4\times {10}^{-12}\text{m}\\ \text{=122}\text{.4}\times {10}^{-12}\text{m}\end{array}$

Substitute $120.0\text{pm}$ for ${\lambda }^{\prime }$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$, $2.998\times {10}^8\text{m/s}$ for $c$ and $120.0°$ for $\theta$ to find the wavelength of scattered photon at $120.0°$.

  $\begin{array}{c}\lambda =120.0\text{pm}\left(\frac{{10}^{-12}\text{m}}{1.0\text{pm}}\right)+\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{\left(9.11\times {10}^{-31}\text{kg}\right)\left(2.998\times {10}^8\text{m/s}\right)}\left(1-\cos \left(120.0°\right)\right)\\ =120.0\times {10}^{-12}\text{m}+3.6\times {10}^{-12}\text{m}\\ \text{=123}\text{.6}\times {10}^{-12}\text{m}\end{array}$

Substitute $120.0\text{pm}$ for ${\lambda }^{\prime }$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$, $2.998\times {10}^8\text{m/s}$ for $c$ and $150.0°$ for $\theta$ to find the wavelength of scattered photon at $150.0°$.

  $\begin{array}{c}\lambda =120.0\text{pm}\left(\frac{{10}^{-12}\text{m}}{1.0\text{pm}}\right)+\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{\left(9.11\times {10}^{-31}\text{kg}\right)\left(2.998\times {10}^8\text{m/s}\right)}\left(1-\cos \left(150.0°\right)\right)\\ =120.0\times {10}^{-12}\text{m}+4.5\times {10}^{-12}\text{m}\\ \text{=124}\text{.5}\times {10}^{-12}\text{m}\end{array}$

Substitute $120.0\text{pm}$ for ${\lambda }^{\prime }$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $9.11\times {10}^{-31}\text{kg}$ for $m_e$, $2.998\times {10}^8\text{m/s}$ for $c$ and $180.0°$ for $\theta$ to find the wavelength of scattered photon at $180.0°$.

  $\begin{array}{c}\lambda =120.0\text{pm}\left(\frac{{10}^{-12}\text{m}}{1.0\text{pm}}\right)+\frac{6.626\times {10}^{-34}\text{J}\cdot \text{s}}{\left(9.11\times {10}^{-31}\text{kg}\right)\left(2.998\times {10}^8\text{m/s}\right)}\left(1-\cos \left(180.0°\right)\right)\\ =120.0\times {10}^{-12}\text{m}+4.8\times {10}^{-12}\text{m}\\ \text{=124}\text{.8}\times {10}^{-12}\text{m}\end{array}$

Therefore, The wavelengths of photons scattered at angles $30.0°$, $60.0°$, $90.0°$ , $120.0°$, $150.0°$, $180.0°$ are $120.3\times {10}^{-12}\text{m}$, $\text{121}\text{.2}\times {10}^{-12}\text{m}$, $\text{122}\text{.4}\times {10}^{-12}\text{m}$, $\text{123}\text{.6}\times {10}^{-12}\text{m}$, $\text{124}\text{.5}\times {10}^{-12}\text{m}$ and $\text{124}\text{.8}\times {10}^{-12}\text{m}$ respectively.

(b)

To determine

The energy of electrons at scattering angles $30.0°$, $60.0°$, $90.0°$ , $120.0°$, $150.0°$, $180.0°$.

Answer to Problem 28P

The energy of electrons at scattering angles $30.0°$, $60.0°$, $90.0°$ , $120.0°$, $150.0°$, $180.0°$ are $27.9\text{eV}$, $104\text{eV}$, $205\text{eV}$, $305\text{eV}$, $376\text{eV}$, $402\text{eV}$ respectively.

Explanation of Solution

The energy of the electrons will be the energy lost from the photons.

Write the expression to find the energy of electrons.

  $\begin{array}{c}{K}_e=\frac{hc}{{\lambda }_0}-\frac{hc}{{\lambda }^{\prime }}\\ =hc\left(\frac{1}{{\lambda }^{\prime }}-\frac{1}{\lambda }\right)\end{array}$

Here, $K_e$ is the energy of electrons, ${\lambda }^{\prime }$ is the wavelength before scattering, $\lambda$ is the wavelength of scattered photon.

Conclusion:

Substitute $120.0\times {10}^{-12}\text{m}$ for ${\lambda }^{\prime }$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $2.998\times {10}^8\text{m/s}$ for $c$ and $120.3\times {10}^{-12}\text{m}$ for $\lambda$ to find the energy of scattered electron when scattering angle is $30.0°$.

  $\begin{array}{c}{K}_e=\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.998\times {10}^8\text{m/s}\right)\left(\frac{1}{120.0\times {10}^{-12}\text{m}}-\frac{1}{120.3\times {10}^{-12}\text{m}}\right)\left(\frac{1.0\text{J}}{1.602\times {10}^{-19}\text{eV}}\right)\\ =27.9\text{eV}\end{array}$

Substitute $120.0\times {10}^{-12}\text{m}$ for ${\lambda }^{\prime }$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $2.998\times {10}^8\text{m/s}$ for $c$ and $\text{121}\text{.2}\times {10}^{-12}\text{m}$ for $\lambda$ to find the energy of scattered electron when scattering angle at $60.0°$.

  $\begin{array}{c}{K}_e=\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.998\times {10}^8\text{m/s}\right)\left(\frac{1}{120.0\times {10}^{-12}\text{m}}-\frac{1}{\text{121}\text{.2}\times {10}^{-12}\text{m}}\right)\left(\frac{1.0\text{J}}{1.602\times {10}^{-19}\text{eV}}\right)\\ =104\text{eV}\end{array}$

Substitute $120.0\times {10}^{-12}\text{m}$ for ${\lambda }^{\prime }$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $2.998\times {10}^8\text{m/s}$ for $c$ and $\text{122}\text{.4}\times {10}^{-12}\text{m}$ for $\lambda$ to find the energy of scattered electron when scattering angle at $90.0°$.

  $\begin{array}{c}{K}_e=\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.998\times {10}^8\text{m/s}\right)\left(\frac{1}{120.0\times {10}^{-12}\text{m}}-\frac{1}{\text{122}\text{.4}\times {10}^{-12}\text{m}}\right)\left(\frac{1.0\text{J}}{1.602\times {10}^{-19}\text{eV}}\right)\\ =205\text{eV}\end{array}$

Substitute $120.0\times {10}^{-12}\text{m}$ for ${\lambda }^{\prime }$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $2.998\times {10}^8\text{m/s}$ for $c$ and $\text{123}\text{.6}\times {10}^{-12}\text{m}$ for $\lambda$ to find the energy of scattered electron when scattering angle at $120.0°$.

  $\begin{array}{c}{K}_e=\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.998\times {10}^8\text{m/s}\right)\left(\frac{1}{120.0\times {10}^{-12}\text{m}}-\frac{1}{\text{123}\text{.6}\times {10}^{-12}\text{m}}\right)\left(\frac{1.0\text{J}}{1.602\times {10}^{-19}\text{eV}}\right)\\ =305\text{eV}\end{array}$

Substitute $120.0\times {10}^{-12}\text{m}$ for ${\lambda }^{\prime }$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $2.998\times {10}^8\text{m/s}$ for $c$ and $\text{124}\text{.5}\times {10}^{-12}\text{m}$ for $\lambda$ to find the energy of scattered electron when scattering angle at $150.0°$.

  $\begin{array}{c}{K}_e=\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.998\times {10}^8\text{m/s}\right)\left(\frac{1}{120.0\times {10}^{-12}\text{m}}-\frac{1}{\text{124}\text{.5}\times {10}^{-12}\text{m}}\right)\left(\frac{1.0\text{J}}{1.602\times {10}^{-19}\text{eV}}\right)\\ =376\text{eV}\end{array}$

Substitute $120.0\times {10}^{-12}\text{m}$ for ${\lambda }^{\prime }$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $2.998\times {10}^8\text{m/s}$ for $c$ and $\text{124}\text{.8}\times {10}^{-12}\text{m}$ for $\lambda$ to find the energy of scattered electron when scattering angle at $180.0°$.

  $\begin{array}{c}{K}_e=\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(2.998\times {10}^8\text{m/s}\right)\left(\frac{1}{120.0\times {10}^{-12}\text{m}}-\frac{1}{\text{124}\text{.8}\times {10}^{-12}\text{m}}\right)\left(\frac{1.0\text{J}}{1.602\times {10}^{-19}\text{eV}}\right)\\ =402\text{eV}\end{array}$

Therefore, The energy of scattered electron when scattering angle is $30.0°$, $60.0°$, $90.0°$, $120.0°$, $150.0°$, $180.0°$ are $27.9\text{eV}$, $104\text{eV}$, $205\text{eV}$, $305\text{eV}$, $376\text{eV}$, $402\text{eV}$ respectively.

(c)

To determine

The scattering angle with which the electron gets the greatest energy.

Answer to Problem 28P

The scattering angle with which the electron gets the greatest energy is $180°$.

Explanation of Solution

When the scattering angle is $180.0°$, the photon strikes on the electron which is at rest with a head on collision. In the head on collision, the photon imparts its momentum to the electron.

After the head on collision the photon scattered straight back and the kinetic energy gained by the initially stationary electron will be the maximum. The kinetic energy gained by the electron at scattering angle $180°$ is $402\text{eV}$.