Question

What is the maximum kinetic energy of electrons knocked out of a thin copper foil by Compton scattering of an incident beam of 17.5 keV x-rays? Assume the work function is negligible.

Short answer

The maximum energy of electron is 1.1 keV.

Step-by-step solution

The given data

Compton scattering of an incident beam of 17.5 keV x-rays is done to knock electrons out a thin copper foil

Formula used

Here, $E\lambda =hc$

Where E is energy

$\lambda$is wavelength

h is Plank’s constant

C is speed of light.

Change in the wavelength

Using $hc=1240\text{\hspace{0.17em}eV}\cdot \text{nm}$

$$\begin{gathered} E\lambda =hc \\ \lambda =\frac{hc}{E} \\ \lambda =\frac{1240\text{\hspace{0.17em}eV}\cdot \text{nm}}{17500\text{\hspace{0.17em}eV}}=0.07086\text{\hspace{0.17em}nm} \end{gathered}$$

Maximum Crompton shift occur for$\phi =180°$ yields

$\begin{array}{c}\Delta \lambda =\left(\frac{hc}{m_e{c}^2}\right)(1-\cos 180°)\\ =\left(\frac{1240\text{\hspace{0.17em}eV}\cdot \text{nm}}{511\times {10}^3\text{\hspace{0.17em}eV}}\right)\left(1-(-1)\right)\\ =0.00485\text{\hspace{0.17em}nm}\end{array}$

New photon energy

New photon wavelength${\lambda }^{\text{'}}$ is

${\lambda }^{\text{'}}=0.07086\text{\hspace{0.17em}nm}+0.00485\text{\hspace{0.17em}nm}=0.0757\text{\hspace{0.17em}nm}$

$\begin{array}{c}{E}^{\text{'}}=\frac{hc}{{\lambda }^{\text{'}}}\\ =\frac{1240\text{\hspace{0.17em}eV}\cdot \text{nm}}{0.0757\text{\hspace{0.17em}nm}}\\ =1.64\times {10}^4\text{\hspace{0.17em}eV}\end{array}$

New photon energy $16.4\text{\hspace{0.17em}keV}$

Kinetic energy of electron

Kinetic energy of electron is

$E^{\text{'}}-E=17.5\text{\hspace{0.17em}keV}-16.4\text{\hspace{0.17em}keV\hspace{0.17em}}=1.1\text{\hspace{0.17em}keV}$

Hence, maximum kinetic energy of electron is $1.1\text{\hspace{0.17em}keV}$