Question

In Fig. 36-47, first-order reflection from the reflection planes shown occurs when an x-ray beam of wavelength$\mathbf{0}\mathbf{.}\mathbf{260}\mathit{n}\mathit{m}$makes an angle$\mathbf{\theta }\mathbf{=}\mathbf{63}\mathbf{.}\mathbf{8}\mathbf{°}$ with the top face of the crystal. What is the unit cell size${\mathbf{a}}_{\mathbf{0}}$?

Short answer

The unit cell distance is $0.570 \text{nm}$.

Step-by-step solution

Bragg’s law

Diffraction is a phenomena in which the interference or the bending of waves occurs around the corners of the opening or obstacle through or on which it strikes.

The diffraction formula is following,

$$\begin{gathered} \mathrm{n\lambda }=2\mathrm{dsin\theta } \\ \mathrm{\lambda }=\text{wavelength} \\ \mathrm{d}=\text{spacing} \\ \mathrm{\theta }=\text{bragg angle} \\ \mathrm{n}=\text{order of diffraction} \end{gathered}$$

Step 2: Identification of the given data

The wavelength is,$\lambda =0.260 \text{nm}$.

The angle is, $63.8°$.

The angle of incidence on the reflection planes is, $\theta =63.8°-45=18.8^0$.

The order of diffraction is,$n=1$ .

 Step 3: The unit cell size

From the given figure it is clear that the relation following will be formed,

$$\begin{gathered} d=\frac{a_0}{\sqrt{2}} \\ {a}_0=\sqrt{2}d \end{gathered}$$

Now from the bragg’s law,

$\sqrt{2}d=\frac{\sqrt{2}n\lambda }{2\sin \theta }$

Substitute all the value in the above equation.

$$\begin{gathered} a_0=\frac{1\times 0.260 \text{nm}}{\sqrt{2}\times \sin 18.8°} \\ {a}_0=0.570 \text{nm} \end{gathered}$$

So the unit cell distance is$0.570\text{nm}$ .