(round off to $2$ decimal places).
Diffraction Data Analysis for BCC Metal
The problem involves finding the atomic radius of a metal with a Body-Centered Cubic (BCC) structure using X-ray diffraction data. We are given the X-ray wavelength ($\lambda$) and the Bragg angle ($\theta$) for the first diffraction peak.
Applying Bragg's Law
Bragg's Law relates the wavelength of incident radiation, the distance between atomic planes, and the angle of diffraction:
$ n\lambda = 2d\sin\theta $
For the first diffraction peak, $n=1$. We are given $\lambda = 0.25$ nm and $\theta = 30^{\circ}$.
Substituting the values:
$ 1 \times (0.25 \text{ nm}) = 2d\sin(30^{\circ}) $
Since $\sin(30^{\circ}) = 0.5$, the equation becomes:
$ 0.25 \text{ nm} = 2d(0.5) $
$ 0.25 \text{ nm} = d $
This value 'd' represents the interplanar spacing ($d_{hkl}$) for the crystal plane causing the first diffraction peak.
BCC Structure Considerations
For a BCC crystal structure, the first diffraction peak corresponds to the (110) plane. The relationship between the interplanar spacing ($d_{110}$) and the lattice parameter ($a$) is:
$ d_{110} = \frac{a}{\sqrt{h^2+k^2+l^2}} = \frac{a}{\sqrt{1^2+1^2+0^2}} = \frac{a}{\sqrt{2}} $
We found $d_{110} = 0.25$ nm. Now we can find the lattice parameter $a$:
$ 0.25 \text{ nm} = \frac{a}{\sqrt{2}} $
$ a = 0.25 \sqrt{2} \text{ nm} $
Calculating Atomic Radius
In a BCC structure, the atoms touch along the body diagonal. The relationship between the lattice parameter ($a$) and the atomic radius ($r$) is:
$ a = \frac{4r}{\sqrt{3}} $
Now, substitute the expression for $a$ and solve for $r$:
$ 0.25 \sqrt{2} \text{ nm} = \frac{4r}{\sqrt{3}} $
Rearranging to solve for $r$:
$ r = \frac{0.25 \sqrt{2} \times \sqrt{3}}{4} \text{ nm} $
$ r = \frac{0.25 \sqrt{6}}{4} \text{ nm} $
Using $\sqrt{6} \approx 2.449$:
$ r \approx \frac{0.25 \times 2.449}{4} \text{ nm} $
$ r \approx \frac{0.61225}{4} \text{ nm} $
$ r \approx 0.153 \text{ nm} $
Rounding to two decimal places, the atomic radius is approximately $0.15$ nm. This value falls within the provided range of 0.13 to 0.17 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)
