Question

X-ray diffraction using a monochromatic radiation of wavelength 0.154 nm is performed on powder samples of metal A (with FCC crystal structure) and metal B (with BCC crystal structure). If the first peak in both the cases occurs at a Bragg angle \( \theta = 20^\circ \), then the value of \(\frac{{Lattice parameter of metal A}}{{Lattice parameter of metal B}} = \ldots\ldots\ldots { (rounded off to two decimal places)}\) .

Hint:

Bragg's law relates the lattice spacing \( d \) to the X-ray wavelength and the Bragg angle. For different crystal structures (FCC vs. BCC), the interplanar spacing \( d \) differs, affecting the calculation of the lattice parameter.

Answer 1

We use Bragg's law for X-ray diffraction:

\[ n \lambda = 2d \sin \theta \] Where \( n \) is the diffraction order, \( \lambda \) is the X-ray wavelength, \( d \) is the interplanar spacing, and \( \theta \) is the Bragg angle.

For the first diffraction peak, \( n = 1 \), so we have:

\[ \lambda = 2d \sin \theta \] The interplanar spacing \( d \) for FCC and BCC crystals is related to the lattice parameter \( a \) by:

- For FCC, \( d = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \), where \( h, k, l \) are the Miller indices.
- For BCC, \( d = \frac{a}{\sqrt{h^2 + k^2 + l^2}} \), where the relationship is different due to the crystal structure.

Since both metals are subjected to the same diffraction conditions and we are comparing their lattice parameters, we can use the relationship for diffraction at \( \theta = 20^\circ \) and the given wavelength \( \lambda = 0.154 \, {nm} \).

Using Bragg's law, we calculate the lattice parameters for both FCC and BCC crystals:

- For FCC: \( a_{{FCC}} \approx 1.20 \, {nm} \)
- For BCC: \( a_{{BCC}} \approx 1.25 \, {nm} \)

Thus, the lattice parameters are between 1.20 and 1.25 nm.