Question

If first-order reflection occurs in a crystal at Bragg angle$3.4°$, at what Bragg angle does second-order reflection occur from the same family of reflecting planes?

Short answer

The angle will be 6.8 degrees.

Step-by-step solution

Introduction of diffraction

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 bragg angle is, ${\theta }_1=3.4°$.

Determination of the Bragg angle at which second order reflection

From the given relationship of the wavelengths the angle will be,

$$\begin{gathered} \frac{2d\sin {\theta }_2}{2d\sin {\theta }_1}=\frac{2\lambda }{1\lambda } \\ \sin {\theta }_2=2\sin {\theta }_1 \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} \sin {\theta }_2=2\times \sin 3.4° \\ \sin {\theta }_2=0.1186 \\ {\theta }_2={\sin }^{-1}0.1186 \\ {\theta }_2=6.8^0 \end{gathered}$$

Hence the angle is,$6.8^0$.