Chapter 26, Problem 58QAP

To determine

The wavelength of x-ray source

Answer to Problem 58QAP

The wavelength of x-ray source is ${\lambda }_i=1.618nm\text{ = 16}\text{.18 Angstrom}$.

Explanation of Solution

Given info:

Scattering angle of the photon is $\theta ={60}^0$ and the speed of electron is $v=4.5\times {10}^5\text{m/s}$. The value of K is $92.13\times {10}^{-21}\text{ J}$

Formula used:

According to Compton scattering the change in wavelength of the incident photon is given by
  $\Delta \lambda =({\lambda }_f-{\lambda }_i)=\frac{h}{m_{e}c}(1-\cos \theta )$

Now the energy of the scattered photon is
$E_f=\frac{hc}{{\lambda }_f}$

Kinetic energy of the scattered electron is given by
  $K=E_i-E_f=hc\left(\frac{{\lambda }_f-{\lambda }_i}{{\lambda }_f{\lambda }_i}\right)=hc\left(\frac{\Delta {\lambda }_{}}{{\lambda }_f{\lambda }_i}\right)=\frac{1}{2}m{v}^2$

Calculation:

Putting $\theta ={60}^0$ in Eq. (1.1) we get
  $\Delta \lambda =({\lambda }_f-{\lambda }_i)=2.43\times {10}^{-12}(1-\cos 60°)\text{ m = 1}\text{.215}\times {\text{10}}^{\text{-12}}\text{m}$

Now putting the given value of K, and calculated $\Delta {\lambda }_{}$ in Eq. (1.3) we get
  $\begin{array}{l}K=hc\left(\frac{\Delta \lambda }{{\lambda }_f{\lambda }_i}\right)=\frac{1}{2}\times 9.1\times {10}^{-31}\times {(}^{4.5}=92.13\times {10}^{-21}\text{ J}\\ \text{or, }{\lambda }_f{\lambda }_i\times {\text{10}}^{\text{-18}}\text{ = }\frac{6.63\times {10}^{-34}\times 3\times {10}^8\times \text{1}\text{.215}\times {\text{10}}^{\text{-12}}}{92.13\times {10}^{-21}}\\ \text{or, }{\lambda }_f{\lambda }_i\text{ = 2}{\text{.62 nm}}^{\text{2}}\\ \text{or, }{\lambda }_f=\frac{\text{2}\text{.62 }}{{\lambda }_i}\text{nm}\end{array}$

Now substituting the value of ${\lambda }_f$ in the expression for $\Delta \lambda =({\lambda }_f-{\lambda }_i)$ we get
  $\begin{array}{l}\Delta \lambda =\left(\frac{2.62}{{\lambda }_i}-{\lambda }_i\right)\text{ nm}=0.001\text{215 nm}\\ \text{or, }2.62-{\lambda }_i{}^2=0.001\text{215}{\lambda }_i\\ \text{or, }{\lambda }_i{}^2+0.001\text{215}{\lambda }_i-2.62=0\\ \end{array}$

Solving this quadratic equation for ${\lambda }_f$ we get
  ${\lambda }_i=\frac{-0.001215\pm \sqrt{{(0.001215)}^2+4\times 1\times (2.62)}}{2\times 1}\text{ nm}$

Now as ${\lambda }_i$ cannot be negative so considering only the positive root of ${\lambda }_i$ we get
  ${\lambda }_i=1.618nm\text{ = 16}\text{.18 Angstrom}$

Conclusion:

So, the initial wavelength of the photon is ${\lambda }_i=1.618nm\text{ = 16}\text{.18 Angstrom}$