Question

Powder X-ray diffraction pattern of a cubic solid with lattice constant \( a \) has the (111) diffraction peak at \( \theta = 30^\circ \). If the lattice expands such that the lattice constant becomes \( 1.25a \), the angle (in degrees) corresponding to the (111) peak changes to \( \sin^{-1} \left( \frac{1}{n} \right) \). The value of \( n \) (rounded off to one decimal place) is _________

Hint:

For changes in lattice constant, the diffraction angle varies inversely with the change in the lattice constant according to Bragg's law.

Answer 1

In X-ray diffraction, the Bragg's law relates the angle \( \theta \) of diffraction to the lattice constant \( a \) of the crystal: \[ n \lambda = 2d \sin \theta \] where:
\( n \) is the diffraction order,
\( \lambda \) is the wavelength of the X-rays,
\( d \) is the interplanar spacing, and
\( \theta \) is the diffraction angle.
For the (111) diffraction peak, the interplanar spacing is given by: \[ d_{111} = \frac{a}{\sqrt{3}} \] Given that the initial diffraction angle \( \theta = 30^\circ \), we can use Bragg's law to express the wavelength \( \lambda \) as: \[ \lambda = 2 d_{111} \sin \theta = 2 \times \frac{a}{\sqrt{3}} \times \sin(30^\circ) = \frac{a}{\sqrt{3}} \] Now, when the lattice constant changes to \( 1.25a \), the new interplanar spacing \( d_{111}' \) becomes: \[ d_{111}' = \frac{1.25a}{\sqrt{3}} \] The diffraction angle \( \theta' \) for this new lattice constant satisfies: \[ \sin \theta' = \frac{\lambda}{2 d_{111}'} = \frac{1}{1.25} = \frac{1}{n} \] Solving for \( n \), we get: \[ n = \frac{1}{\sin \theta'} = 2.5 \] Thus, the value of \( n \) is 2.5.