Calculating FCC Lattice Parameter using XRD Data

This problem requires calculating the lattice parameter ('a') of a face-centered cubic (fcc) metal using information from an X-ray diffraction (XRD) experiment. We can use Bragg's Law and the formula for interplanar spacing in cubic crystals.

Applying Bragg's Law and Interplanar Spacing Formula

1. Identify Given Parameters:
   • Diffraction angle: $2\theta = 78^\circ$
   • Plane indices: {311}
   • Wavelength: $\lambda = 0.154$ nm
   • Crystal structure: fcc
2. Calculate the angle $\theta$:

   $\theta = \frac{2\theta}{2} = \frac{78^\circ}{2} = 39^\circ$

3. Determine the sum of squares for the plane indices:

   For {311} planes in an fcc lattice, the indices are $h=3, k=1, l=1$. The sum of squares is:

   $h^2 + k^2 + l^2 = 3^2 + 1^2 + 1^2 = 9 + 1 + 1 = 11$

4. Use Bragg's Law to find interplanar spacing ($d$):

   Bragg's Law is $n\lambda = 2d \sin\theta$. Assuming the first-order peak ($n=1$):

   $\lambda = 2d \sin\theta$

   Rearranging for $d$:

   $d = \frac{\lambda}{2 \sin\theta}$

   Substitute the values:

   $d = \frac{0.154 \text{ nm}}{2 \times \sin(39^\circ)}$

   Using $\sin(39^\circ) \approx 0.6293$:

   $d \approx \frac{0.154 \text{ nm}}{2 \times 0.6293} \approx \frac{0.154}{1.2586} \approx 0.12236$ nm

5. Relate interplanar spacing to lattice parameter for fcc:

   The formula for interplanar spacing ($d_{hkl}$) in a cubic system is:

   $d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$

   Rearranging to solve for the lattice parameter 'a':

   $a = d_{hkl} \times \sqrt{h^2 + k^2 + l^2}$

6. Calculate the lattice parameter ('a'):

   Substitute the calculated $d$ and the sum of squares:

   $a \approx 0.12236 \text{ nm} \times \sqrt{11}$

   Using $\sqrt{11} \approx 3.3166$:

   $a \approx 0.12236 \text{ nm} \times 3.3166 \approx 0.4058$ nm

7. Conclusion:

   The calculated lattice parameter is approximately 0.4058 nm, which is closest to the option 0.4 nm.