Question

As shown in the figure, X-ray diffraction pattern is obtained from a diatomic chain of atoms P and Q. The diffraction condition is given by \( a \cos \theta = n\lambda \), where \( n \) is the order of the diffraction peak. Here, \( a \) is the lattice constant and \( \lambda \) is the wavelength of the X-rays. Assume that atomic form factors and resolution of the instrument do not depend on \( \theta \). Then, the intensity of the diffraction peaks is 

• A. lower for even values of \( n \), when compared to odd values of \( n \)
• B. lower for odd values of \( n \), when compared to even values of \( n \)
• C. zero for odd values of \( n \)
• D. zero for even values of \( n \)

Hint:

In diatomic lattices, the intensity of diffraction peaks for even \( n \) is typically lower than for odd \( n \) due to destructive interference between the contributions from the two types of atoms.

Answer 1

The Correct Option is B

To solve this problem, we need to analyze the diffraction pattern from a diatomic chain of atoms using the given condition:

\(a \cos \theta = n\lambda\)

where \(a\) is the lattice constant, \(n\) is the order of the diffraction peak, \(\theta\) is the angle, and \(\lambda\) is the wavelength of the X-rays.

For a diatomic chain composed of two different atoms, P and Q, the intensity of the diffraction peaks depends on the basis of the lattice. In general, for such a diatomic chain, the diffraction condition will lead to different intensities for different orders of peaks. This is because the scattering factors of the atoms (P and Q) can lead to constructive or destructive interference.

Let's analyze the intensity pattern:

1. For even orders of \(n\) (e.g., \(n=2, 4, 6, \ldots\)), the diffraction peaks tend to constructively interfere, leading to higher intensity.
2. For odd orders of \(n\) (e.g., \(n=1, 3, 5, \ldots\)), the diffraction peaks tend to destructively interfere. This happens because the alternating positions of P and Q atoms cause cancellation of scattering amplitudes, resulting in lower intensity.

This means that the intensity of diffraction peaks is indeed lower for odd values of \(n\) compared to even values of \(n\).

Hence, the correct answer is:

lower for odd values of \(n\), when compared to even values of \(n\)