Question

From equations (3-4) to (3-6) obtain equations (3.8). It is easiest to start by eliminating φ between equations (3-4) and (3-5) using cos²φ+sin²φ = 1 The electron speed u may then be eliminated between the remaining equations.

Short answer

Using the given equations, the following can be obtained is $\lambda \text{'}-\lambda =\frac{h}{m_{e}c}\left(1-\cos \theta \right)$.

Step-by-step solution

Given data

$$\begin{gathered} \frac{h}{\lambda }=\frac{h}{\lambda \text{'}}\cos \theta +{\gamma }_u{m}_{e}u\cos \varphi \dots ..\left(1\right) \\ 0=\frac{h}{\lambda \text{'}}\sin \theta -{\gamma }_u{m}_{e}u\sin \varphi \dots ..\left(2\right) \\ \frac{hc}{\lambda }-m_e{c}^2=\frac{hc}{\lambda \text{'}}+{\gamma }_u{m}_e{c}^2\dots ..\left(3\right) \end{gathered}$$

Concept  used

Einstein's mass-energy equivalence relation can be expressed as,

E = mc²

Use the equations and solve

Rearranging the above equations (1 and 2) will give:

$$\begin{gathered} {\gamma }_u{m}_{e}u\cos \varphi =\frac{h}{\lambda }-\frac{h}{\lambda \text{'}}\cos \theta \\ {\gamma }_u{m}_{e}u\sin \varphi =\frac{h}{\lambda \text{'}}\sin \theta \end{gathered}$$

Squaring both will give:

$$\begin{gathered} {\left({\gamma }_u{m}_{e}u\right)}^2{\cos }^2\varphi ={\left(\frac{h}{\lambda }-\frac{h}{\lambda \text{'}}\cos \theta \right)}^2 \\ {\left({\gamma }_u{m}_{e}u\right)}^2{\sin }^2\varphi ={\left(\frac{h}{\lambda \text{'}}\sin \theta \right)}^2 \end{gathered}$$

Adding both equations will lead to:

${\left({\gamma }_u{m}_{e}u\right)}^2={\left(\frac{h}{\lambda }\right)}^2-\frac{2h^2}{\lambda \lambda \text{'}}\cos \theta +{\left(\frac{h}{\lambda \text{'}}\right)}^2$…..(4)

The Square of equation 3 will give

${\left({\gamma }_u{m}_{e}u\right)}^2={\left(\frac{h}{\lambda }\right)}^2+{\left(\frac{h}{\lambda }\right)}^2-\frac{2h^2}{\lambda \lambda \text{'}}+2m_{e}c\left(\frac{h}{\lambda }-\frac{h}{\lambda \text{'}}\right)+m_e^2{c}^2$…..(5)

Subtract equation 5 from equation 4

Simplify further,

$\begin{array}{rcl}{m}_e^2\left({\gamma }_u^2{c}^2-{\gamma }_u^2{c}^2\right)& =& -\frac{2h^2}{\lambda \lambda \text{'}}\left(1-\cos \theta \right)+2m_{e}c\left(\frac{h}{\lambda \text{'}}-\frac{h}{\lambda }\right)+{\left(m_{e}c\right)}^2\\ \frac{2h^2}{\lambda \lambda \text{'}}\left(1-\cos \theta \right)& =& 2m_{e}c\left(\frac{h}{\lambda \text{'}}-\frac{h}{\lambda }\right)\\ \frac{h}{\lambda \lambda \text{'}}\left(1-\cos \theta \right)& =& m_{e}c\left(\frac{1}{\lambda \text{'}}-\frac{1}{\lambda }\right)\\ \lambda -\lambda \text{'}& =& \frac{h}{m_{e}c}\left(1-\cos \theta \right)\end{array}$