Calculating FCC Lattice Parameter using X-ray Diffraction

This solution outlines the steps to determine the lattice parameter of an elemental metal with a Face-Centered Cubic (FCC) crystal structure using X-ray diffraction data.

Applying Bragg's Law

Bragg's Law relates the wavelength of X-rays ($\lambda$), the distance between atomic planes ($d$), the Bragg angle ($\theta$), and the order of diffraction ($n$):

$n\lambda = 2d\sin\theta$

For the first peak ($n=1$), we can rearrange the formula to find the interplanar spacing ($d$):

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

Given:

• Wavelength ($\lambda$) = $0.1543$ nm
• Bragg angle ($\theta$) = $24.65^\circ$
• Order ($n$) = 1

First, calculate $\sin\theta$:

$\sin(24.65^\circ) \approx 0.4171$

Now, calculate $d$:

$d = \frac{1 \times 0.1543 \text{ nm}}{2 \times 0.4171} \approx \frac{0.1543 \text{ nm}}{0.8342} \approx 0.18497 \text{ nm}$

Relating Interplanar Spacing to Lattice Parameter for FCC

For an FCC crystal structure, the interplanar spacing ($d$) for a set of Miller indices ($hkl$) is given by:

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

where $a$ is the lattice parameter.

The first diffraction peak in an FCC pattern corresponds to the (111) planes because they represent the smallest allowed value for $\sqrt{h^2+k^2+l^2}$.

Therefore, for the (111) plane:

$d_{(111)} = \frac{a}{\sqrt{1^2+1^2+1^2}} = \frac{a}{\sqrt{3}}$

Calculating the Lattice Parameter

Rearrange the formula to solve for $a$:

$a = d_{(111)}\sqrt{3}$

Substitute the calculated value of $d$:

$a \approx 0.18497 \text{ nm} \times \sqrt{3}$

$a \approx 0.18497 \text{ nm} \times 1.732 \approx 0.3203 \text{ nm}$

Rounding to two decimal places, the lattice parameter is $0.32$ nm.