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

Constructive interference from parallel crystal planes obeys the Bragg condition $2d\sin\theta = n\lambda$, with $n=1$ for the first observed reflection.

Here the path difference between rays scattered off successive planes equals one wavelength. Solving for $d$ gives $d = \dfrac{\lambda}{2\sin\theta}$ for first order.

Substitute $\lambda = 0.16\ \text{nm}$ and $\theta = 30^\circ$. Since $\sin 30^\circ = 1/2$, the factor $2\sin\theta = 2 \times 0.5 = 1$.

Therefore $d = \dfrac{0.16\ \text{nm}}{1} = 0.16\ \text{nm}$. The spacing happens to equal the wavelength here because $2\sin\theta = 1$ exactly at this angle. \[\boxed{d = 0.16\ \text{nm}}\]