Question

Prove that fur any sine function $\mathit{s}\mathit{i}\mathit{n}\left(kx+\varphi \right)$of wavelength shorter than 2a, where ais the atomic spacing. there is a sine function with a wavelength longer than 2a that has the same values at the points x = a , 2a , 3a . and so on. (Note: It is probably easier to work with wave number than with wavelength. We sick to show that for every wave number greater than there is an equivalent less than $\mathit{\pi }\mathbf{/}\mathit{a}$.)

Short answer

It is proved that there is a sine function with a wavelength less than 2a.

Step-by-step solution

Sum of angle and expression for θ:

The atomic spacing is a.

The expression of sum of angles is given by,

${\mathit{\theta }}_{\mathbf{1}}\mathbf{+}{\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathit{n}\mathit{\pi }$

The expression for is given by,

$\mathit{\theta }\mathbf{=}\mathit{k}\mathit{x}\mathbf{+}\mathit{\varphi }$

Determine sum of angles:

The expression of sum of angles is calculated as,

$$\begin{gathered} {\theta }_1+{\theta }_2=n\pi n \\ \left(k_{1}x+\varphi \right)+\left(k_{2}x+\varphi \right)=n\pi n \\ {k}_{1}x+k_{2}x+2\varphi =n\pi n..........\left(1\right) \end{gathered}$$

For

$\begin{array}{rcl}\left(\frac{\pi }{a}\right)\left(na\right)+\varphi & =& \frac{n\pi }{2}\varphi \\ & =& \frac{n\pi }{2}-n\pi \\ & =& -\frac{n\pi }{2}\end{array}$

solve further:

Substitute $\begin{array}{rcl}& & -\frac{n\pi }{2}\end{array}$for $\varphi$in equation (1).

$\begin{array}{rcl}\left(k_1+k_2\right)x+2\left(-\frac{n\pi }{2}\right)& =& n\mathrm{\pi }\\ \left({\mathrm{k}}_1+{\mathrm{k}}_2\right)\left(\mathrm{na}\right)& =& 2\mathrm{n\pi }\\ {\mathrm{k}}_1+{\mathrm{k}}_2& =& \frac{2\mathrm{\pi }}{\mathrm{a}}\\ {\mathrm{k}}_1& =& \frac{2\mathrm{\pi }}{\mathrm{a}}-{\mathrm{k}}_2\end{array}$

Now since,

$k_1>\frac{\mathrm{\pi }}{a},\frac{2\mathrm{\pi }}{a}-k_2>\frac{\mathrm{\pi }}{a},k_2<\frac{\mathrm{\pi }}{a}$

Therefore, it is proved that there is sine function with wavelength less than 2a.