Question

A metal having body-centered cubic structure is analyzed through X-ray diffraction using monochromatic X-ray of wavelength 0.154 nm. The diffraction angle (\(2\theta\)) corresponding to \(\{2\,0\,0\}\) plane is \(60^{\circ}\) (for first order reflection).
The atomic radius of this element (rounded off to three decimal places) is ______ nm.

Hint:

Find \(d_{200}\) from Bragg's law, then \(a=2d_{200}\) for the \(\{200\}\) plane, then use \(R=a\sqrt3/4\) for BCC.

Answer 1

Correct Answer: 0.132

Step 1: Apply Bragg's law to find the interplanar spacing.
Bragg's law for diffraction is
\[ n\lambda=2d\sin\theta \]
Here the order of reflection is \(n=1\), the wavelength is \(\lambda=0.154\ \text{nm}\), and the diffraction angle given is \(2\theta=60^{\circ}\), so \(\theta=30^{\circ}\).

Step 2: Solve for the interplanar spacing \(d_{200}\).
\[ d_{200}=\frac{n\lambda}{2\sin\theta}=\frac{1\times0.154}{2\sin(30^{\circ})} \]
Since \(\sin(30^{\circ})=0.5\),
\[ d_{200}=\frac{0.154}{2\times0.5}=\frac{0.154}{1}=0.154\ \text{nm} \]

Step 3: Relate the interplanar spacing to the lattice parameter.
For a cubic crystal, the general relation between the interplanar spacing and Miller indices \((hkl)\) is
\[ d_{hkl}=\frac{a}{\sqrt{h^2+k^2+l^2}} \]
For the \(\{2\,0\,0\}\) plane, \(h=2,\ k=0,\ l=0\), so
\[ \sqrt{h^2+k^2+l^2}=\sqrt{4}=2 \]

Step 4: Solve for the lattice parameter \(a\).
\[ a=d_{200}\times2=0.154\times2=0.308\ \text{nm} \]

Step 5: Use the BCC packing relation to find the atomic radius.
In a body-centered cubic structure, atoms touch along the body diagonal, which has length \(a\sqrt3\) and is spanned by \(4\) atomic radii:
\[ 4R=a\sqrt3 \quad\Rightarrow\quad R=\frac{a\sqrt3}{4} \]
Substituting \(a=0.308\ \text{nm}\):
\[ R=\frac{0.308\times1.7321}{4}=\frac{0.5335}{4} \]
\[ R\approx0.1334\ \text{nm} \]

Final Answer:
The atomic radius, rounded to three decimal places, is
\[ \boxed{0.133\ \text{nm}} \]