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.
BCC Structure Analysis via X-ray Diffraction
This solution explains how to calculate the atomic radius of a metal with a Body-Centered Cubic (BCC) crystal structure using X-ray diffraction measurements.
Key Formulas
- Bragg's Law: This fundamental law relates the wavelength of X-rays to the spacing between crystal planes and the angle of diffraction.
$ n\lambda = 2d\sin\theta $
Here, n is the order of reflection, λ denotes the X-ray wavelength, d represents the interplanar spacing, and θ is the Bragg angle. - Interplanar Spacing (d) for BCC: For a BCC lattice, the distance between planes with Miller indices {hkl} depends on the lattice parameter a.
$ d_{hkl} = \frac{a}{\sqrt{h^2+k^2+l^2}} $
Specifically for the {200} planes:$ d_{200} = \frac{a}{\sqrt{2^2+0^2+0^2}} = \frac{a}{2} $
- Lattice Parameter (a) vs. Atomic Radius (r) for BCC: In a BCC structure, atoms touch along the body diagonal.
$ a = \frac{4r}{\sqrt{3}} $
Calculation Steps
- Find the Bragg Angle (θ):
The problem provides the diffraction angle $2\theta = 60^\circ$. The Bragg angle is half of this:
$ \theta = \frac{60^\circ}{2} = 30^\circ $
- Calculate Interplanar Spacing (d200):
Apply Bragg's Law using the given values: $n=1$, $\lambda=0.154$ nm, and $\theta=30^\circ$.
$ 1 \times 0.154 \text{ nm} = 2 \times d_{200} \times \sin(30^\circ) $
Since $\sin(30^\circ) = 0.5$:$ 0.154 \text{ nm} = 2 \times d_{200} \times 0.5 $
$ 0.154 \text{ nm} = d_{200} $
- Determine the Lattice Parameter (a):
Use the relation $d_{200} = \frac{a}{2}$ for BCC:
$ a = 2 \times d_{200} = 2 \times 0.154 \text{ nm} = 0.308 \text{ nm} $
- Calculate the Atomic Radius (r):
Rearrange the BCC formula $a = \frac{4r}{\sqrt{3}}$ to solve for r:
$ r = \frac{a\sqrt{3}}{4} $
Substitute the value of a:$ r = \frac{0.308 \text{ nm} \times \sqrt{3}}{4} $
$ r \approx \frac{0.308 \times 1.73205}{4} \approx \frac{0.5335394}{4} \approx 0.13338 \text{ nm} $
- Round the Result:
Rounding the calculated atomic radius to three decimal places:
$ r \approx 0.133 \text{ nm} $
The atomic radius of the element is approximately 0.133 nm.
Important Questions from Crystal Structure Density Atomic Packing Factor
- Match the crystal systems in Column I with the corresponding axial lengths (a, b, c) and interaxial angles ($\alpha$, $\beta$, $\gamma$) provided in Column II
Column I Column II (P) Tetragonal (1) $a \neq b \neq c$, $\alpha = \beta = \gamma = 90^\circ$ (Q) Rhombohedral (2) $a = b \neq c$, $\alpha = \beta = \gamma = 90^\circ$ (R) Orthorhombic (3) $a \neq b \neq c$, $\alpha = \gamma = 90^\circ \neq \beta$ (S) Monoclinic (4) $a = b = c$, $\alpha = \beta = \gamma \neq 90^\circ$ - The coordination number for an octahedral site in pure copper is __________.
The lattice parameter of face-centered cubic iron ($\gamma$-Fe) is 0.3571 nm. The radius (in nm) of the octahedral void in $\gamma$-Fe is _______________
- For a bcc metal the ratio of the surface energy per unit area of the (100) plane to that of the (110) plane is ________
- Pure iron transforms from body centered cubic (BCC) to face centered cubic (FCC) crystal structure at $912 \text{ °C}$. If the lattice parameter of the BCC phase is $0.293 \text{ nm}$ and that of the FCC phase is $0.363 \text{ nm}$, the associated volume change is ________ (in % to one decimal place)
