Question

An X-ray beam of wavelength \(0.16\ \text{nm}\) is incident on a set of planes of a certain crystal. The first Bragg reflection is observed for an incidence angle of \(30^\circ\). What is the corresponding interplanar spacing?

• A. \(0.16\ \text{nm}\)
• B. \(0.67\ \text{nm}\)
• C. \(1.02\ \text{nm}\)
• D. \(0.89\ \text{nm}\)

Hint:

Apply Bragg's law \(n\lambda = 2d\sin\theta\) with \(n=1\) and \(\sin 30^\circ = 0.5\).

Answer 1

The Correct Option is A

Step 1: Bragg's law for X-ray diffraction is \(n\lambda = 2d\sin\theta\), where \(\theta\) is the glancing (Bragg) angle, \(d\) is the interplanar spacing, and \(n\) is the order.

Step 2: The first reflection means \(n = 1\). Given \(\lambda = 0.16\ \text{nm}\) and \(\theta = 30^\circ\), with \(\sin 30^\circ = 0.5\).

Step 3: Rearrange for the spacing: \[d = \frac{n\lambda}{2\sin\theta} = \frac{1 \times 0.16}{2 \times 0.5}\ \text{nm}.\]

Step 4: The denominator is \(2 \times 0.5 = 1\), so \[d = \frac{0.16}{1} = 0.16\ \text{nm}.\] \[\boxed{d = 0.16\ \text{nm}}\]