Question

Derive Eq. 38-11, the equation for the Compton shift, from Eqs. 38-8, 38-9, and 38-10 by eliminating v and $\mathit{\theta }$.

Short answer

The equation 38-11 is derived as follows:

${\lambda }^{\text{'}}-\lambda =\Delta \lambda =\frac{h}{mc}\left(1-\cos \varphi \right)$

Step-by-step solution

Introduction

Consider the equation 38-8

$\frac{h}{m}\left(\frac{1}{\lambda }-\frac{1}{\lambda \text{'}}\right)+1=\frac{v}{\sqrt{1-{\left({\displaystyle \raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}\right)}^2}}$

Consider the equation 38-9

$\frac{h}{m\lambda }-\frac{h}{m\lambda \text{'}}\cos \varphi =\frac{v}{\sqrt{1-{\left({\displaystyle \raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$$}\right.}\right)}^2}}\cos \theta$

Consider the equation 38-10

$\frac{h}{m\lambda \text{'}}\sin \varphi =\frac{v}{\sqrt{1-{\left({\displaystyle \raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}\right)}^2}}\sin \theta$

Determine the derivation

On squaring and adding the Equation 38-9 and Equation 38-10. From the equation ${\sin }^2\theta +{\cos }^2\theta =1$solve as:

$$\begin{gathered} {\left(\frac{h}{m}\right)}^2\left[{\left(\frac{1}{\lambda }-\frac{1}{\lambda \text{'}}\cos \varphi \right)}^2+{\left(\frac{1}{\lambda \text{'}}\sin \varphi \right)}^2\right]=\frac{v^2}{1-{\left({\displaystyle \raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}\right)}^2} \\ {\left(\frac{h}{m}\right)}^2\left[{\left(\frac{1}{\lambda }-\frac{1}{\lambda \text{'}}\cos \varphi \right)}^2+{\left(\frac{1}{\lambda \text{'}}\sin \varphi \right)}^2\right]=-c^2\left(1-\frac{1}{1-{\left({\displaystyle \raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}\right)}^2}\right) \\ {\left(\frac{h}{mc}\right)}^2\left[{\left(\frac{1}{\lambda }-\frac{1}{\lambda \text{'}}\cos \varphi \right)}^2+{\left(\frac{1}{\lambda \text{'}}\sin \varphi \right)}^2\right]+1=\left(\frac{1}{1-{\left({\displaystyle \raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}\right)}^2}\right) \end{gathered}$$

Now squaring the Equation 38-8, and on comparing the above equation.

$$\begin{gathered} {\left(\frac{h}{mc}\left(\frac{1}{\lambda }-\frac{1}{\lambda \text{'}}\right)+1\right)}^2=\left(\frac{v^2}{1-{\left({\displaystyle \raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$c$}\right.}\right)}^2}\right) \\ {\left(\frac{h}{mc}\left(\frac{1}{\lambda }-\frac{1}{\lambda \text{'}}\right)+1\right)}^2={\left(\frac{h}{mc}\right)}^2\left[{\left(\frac{1}{\lambda }-\frac{1}{\lambda \text{'}}\cos \varphi \right)}^2+{\left(\frac{1}{\lambda \text{'}}\sin \varphi \right)}^2\right]+1 \end{gathered}$$

On solving the above equation we will get this equation that is known as Equation 38-11

$\lambda \text{'}-\lambda =∆\lambda =\frac{h}{mc}\left(1-\cos \varphi \right)$