Question

Diffraction pattern of a polycrystalline BCC metal is obtained using monochromatic X-rays of wavelength 0.25 nm. If the first peak occurs at Bragg angle (\(\theta\)) of 30\(^{\circ}\), then the radius of the metal atom in nm is ................... (round off to 2 decimal places).

Hint:

Memorize the first diffraction peaks for common crystal structures: - {FCC:} (111) - {BCC:} (110) - {Simple Cubic:} (100) This will save you time in identifying the correct (hkl) for the first peak.

Answer 1

Step 1: Understanding the Concept:
This problem combines Bragg's Law of diffraction with the crystallography of a Body-Centered Cubic (BCC) metal. We need to identify the Miller indices of the first diffraction peak for BCC, use Bragg's Law to find the lattice parameter, and then relate the lattice parameter to the atomic radius for the BCC structure.
Step 2: Key Formula or Approach:
1. Bragg's Law: \( n\lambda = 2d_{hkl}\sin\theta \). For the first peak, we use \(n=1\).
2. BCC Selection Rule: Diffraction peaks occur only for planes (hkl) where the sum \(h+k+l\) is an even number.
3. First Peak for BCC: The first allowed reflection (lowest \(\theta\), largest \(d\)-spacing) is from the \{110\} family of planes.
4. Interplanar Spacing (Cubic): \( d_{hkl} = \frac{a}{\sqrt{h^2+k^2+l^2}} \). For (110), \( d_{110} = \frac{a}{\sqrt{1^2+1^2+0^2}} = \frac{a}{\sqrt{2}} \).
5. Atomic Radius (BCC): The atoms touch along the body diagonal, so \( \sqrt{3}a = 4R \), or \( R = \frac{\sqrt{3}a}{4} \).
Step 3: Detailed Calculation:
1. Find the interplanar spacing \(d_{110}\):
Using Bragg's law with \(n=1\), \(\lambda = 0.25\) nm, and \(\theta = 30^{\circ}\):
\[ d_{110} = \frac{\lambda}{2\sin\theta} = \frac{0.25 \text{ nm}}{2\sin(30^{\circ})} = \frac{0.25 \text{ nm}}{2 \times 0.5} = 0.25 \text{ nm} \] 2. Find the lattice parameter \(a\):
We know that \(d_{110} = \frac{a}{\sqrt{2}}\).
\[ 0.25 \text{ nm} = \frac{a}{\sqrt{2}} \implies a = 0.25 \times \sqrt{2} \approx 0.35355 \text{ nm} \] 3. Find the atomic radius \(R\):
Using the relationship for BCC structures:
\[ R = \frac{\sqrt{3}a}{4} = \frac{\sqrt{3} \times 0.35355}{4} \approx \frac{1.732 \times 0.35355}{4} \approx \frac{0.61237}{4} \approx 0.15309 \text{ nm} \] 4. Round to 2 decimal places: \[ R \approx 0.15 \text{ nm} \] Step 4: Final Answer:
The radius of the metal atom is 0.15 nm.