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.
Show Hint
Find \(d_{200}\) from Bragg's law, then \(a=2d_{200}\) for the \(\{200\}\) plane, then use \(R=a\sqrt3/4\) for BCC.
Step 1: Combine all three relations into one formula before putting in numbers.
Three facts are needed: Bragg's law $n\lambda=2d\sin\theta$, the cubic spacing rule $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$, and the BCC touching condition $4R=a\sqrt3$. Chain them together algebraically first.
Step 2: Express $a$ in terms of $d_{200}$.
For $\{200\}$, $\sqrt{h^2+k^2+l^2}=2$, so $a=2d_{200}$. And from Bragg's law, $d_{200}=\dfrac{n\lambda}{2\sin\theta}$. So
\[ a=2\cdot\frac{n\lambda}{2\sin\theta}=\frac{n\lambda}{\sin\theta} \]
Step 3: Substitute $a$ into the BCC radius formula.
\[ R=\frac{a\sqrt3}{4}=\frac{\sqrt3}{4}\cdot\frac{n\lambda}{\sin\theta} \]
This single formula now links the radius directly to the measured angle and wavelength, without needing $d$ or $a$ as separate intermediate numbers.
Step 4: Plug in the numbers only once.
With $n=1$, $\lambda=0.154\ \text{nm}$, and $\theta=30^{\circ}$ so $\sin\theta=0.5$:
\[ R=\frac{\sqrt3}{4}\times\frac{1\times0.154}{0.5}=\frac{1.7321}{4}\times0.308 \]
Step 5: Finish the arithmetic.
\[ R=0.4330\times0.308\approx0.1334\ \text{nm} \]
Rounded to three decimal places, this gives the same result as computing $d$ and $a$ separately.
\[ \boxed{R\approx0.133\ \text{nm}} \]
