The metal iridium has an FCC crystal structure. If the angle of diffraction for the (220) set of planes occurs at 69.22 (first-order reflection) when monochromatic x-radiation having a wavelength of 0.1542 nm is used, compute 
(a) the interplanar spacing for this set of planes, and (b) the atomic radius for an iridium atom. 


To solve this problem, we can use Bragg's Law and the geometry of the FCC (face-centered cubic) crystal structure. Here's the step-by-step approach:

### (a) Compute the interplanar spacing for the (220) set of planes

We will use **Bragg’s Law**:

\[
n\lambda = 2d\sin\theta
\]

Where:
- \( n \) is the order of reflection (in this case, \( n = 1 \)),
- \( \lambda = 0.1542 \, \text{nm} \) (wavelength of the x-rays),
- \( \theta = 69.22^\circ \) (the diffraction angle),
- \( d \) is the interplanar spacing, which we want to compute.

We can rearrange Bragg’s Law to solve for \( d \):

\[
d = \frac{n\lambda}{2\sin\theta}
\]

Now, substituting the known values:

\[
d = \frac{1 \times 0.1542 \, \text{nm}}{2 \times \sin(69.22^\circ)}
\]

Let’s compute \( d \).The interplanar spacing for the (220) set of planes is approximately \( d = 0.0825 \, \text{nm} \).

### (b) Compute the atomic radius for an iridium atom

For a face-centered cubic (FCC) crystal structure, the relationship between the interplanar spacing \( d \) for a particular set of planes \( (hkl) \), the lattice parameter \( a \), and the Miller indices \( h, k, l \) is given by:

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

For the \( (220) \) planes, \( h = 2 \), \( k = 2 \), and \( l = 0 \), so:

\[
d_{220} = \frac{a}{\sqrt{2^2 + 2^2 + 0^2}} = \frac{a}{\sqrt{8}} = \frac{a}{2\sqrt{2}}
\]

Rearranging to solve for \( a \):

\[
a = 2\sqrt{2} \, d
\]

Now we can compute the lattice parameter \( a \) using the value of \( d \).

Afterward, for an FCC structure, the atomic radius \( r \) is related to the lattice parameter \( a \) by the formula:

\[
r = \frac{a}{2\sqrt{2}}
\]

Let's calculate the atomic radius \( r \).The atomic radius of an iridium atom is approximately \( r = 0.0825 \, \text{nm} \), and the lattice parameter \( a = 0.2332 \, \text{nm} \).

Would you like more details or clarification on any step?

Here are some related questions to explore:
1. What are the general characteristics of FCC crystal structures?
2. How does the atomic radius affect the physical properties of metals like iridium?
3. How can the Bragg's Law be used in determining the crystal structure of other metals?
4. What are Miller indices and how do they relate to crystal planes?
5. How do x-rays interact with crystal structures in diffraction experiments?

**Tip**: Understanding crystal structures, such as FCC, BCC, and HCP, is essential in material science for predicting mechanical properties.

The metal iridium has an FCC crystal structure. If the angle of diffraction for the (220) set of planes occurs at 69.22° (first-order reflection) when monochromatic x-radiation having a wavelength of 0.1542 nm is used, compute (a) the interplanar spacing for this set of planes, and (b) the atomic radius for an iridium atom.

Learn how to compute the interplanar spacing and atomic radius for FCC iridium. Using Bragg's Law and X-ray diffraction data with a 69.22° diffraction angle and 0.1542 nm wavelength, we find the interplanar spacing and atomic radius for an iridium atom. Suitable for undergraduate materials science or physics students.

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