Question

X-rays of wavelength $10\mathring{\text{A}}$ is incident on a crystal plane at an angle of ${60}^{\circ }$ from the normal. What should be the distance between the planes, so that the X-rays undergo constructive interference of second order?

• A. $10\mathring{\text{A}}$
• B. $\frac{20}{\sqrt{3}}\mathring{\text{A}}$
• C. $\frac{10}{\sqrt{3}}\mathring{\text{A}}$
• D. $20\mathring{\text{A}}$

Answer 1

The correct option is D $20\mathring{\text{A}}$

Using Bragg's law, for maxima,

$2d\sin \theta =n\lambda$

Here, $n=2$ for second maxima and $\theta$ is the glancing angle.
$\Rightarrow \theta ={90}^{\circ }-{60}^{\circ }={30}^{\circ }$

$2d\sin {30}^{\circ }=2\times 10$

$\therefore d=20\mathring{\text{A}}$

Hence, $\left(D\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question?}\\ \text{Application of Bragg's law single step calculation.}\end{array}$ Key concept: Formula is extremely useful to calculate wavelength of x-rays and also to determine the spacing between the crystal layers and hence the type of structure.