Question

X-ray diffraction using a monochromatic radiation of wavelength 0.154 nm is performed on powder samples of metal A (FCC crystal structure) and metal B (BCC crystal structure). If the first peak in both the cases occurs at a Bragg angle equal to 20 degrees, then the value of lattice parameter of metal A divided by lattice parameter of metal B is ........... (rounded off to two decimal places).

Hint:

For FCC the first diffraction peak comes from plane (111) and for BCC it comes from plane (110). Use Bragg's law to calculate interplanar spacing, then relate with lattice parameter.

Answer 1

Step 1: Recall Bragg's law.
\[ n\lambda = 2d\sin\theta \] For first order reflection, $n=1$. Hence: \[ d = \frac{\lambda}{2\sin\theta} \] Step 2: Relation between lattice parameter and d-spacing.
- For FCC, first allowed reflection is from plane (111). \[ d_{111} = \frac{a}{\sqrt{3}} \] - For BCC, first allowed reflection is from plane (110). \[ d_{110} = \frac{a}{\sqrt{2}} \] Step 3: Equating using Bragg's law.
Since both metals have the same Bragg angle, the interplanar spacing values are equal to $\frac{\lambda}{2\sin\theta}$. For FCC metal A: \[ d_{111} = \frac{a_A}{\sqrt{3}} \] For BCC metal B: \[ d_{110} = \frac{a_B}{\sqrt{2}} \] Step 4: Ratio of lattice parameters.
Equating $d_{111}$ and $d_{110}$ to the same Bragg law expression: \[ \frac{a_A}{\sqrt{3}} = \frac{a_B}{\sqrt{2}} \] Therefore: \[ \frac{a_A}{a_B} = \frac{\sqrt{3}}{\sqrt{2}} \approx 1.22 \] But when computed with $\theta = 20$ and $\lambda = 0.154$ nm: \[ d = \frac{0.154}{2 \sin 20^\circ} = \frac{0.154}{0.684} = 0.225 \, \text{nm} \] Now, \[ a_A = d \sqrt{3} = 0.225 \times 1.732 = 0.390 \, \text{nm} \] \[ a_B = d \sqrt{2} = 0.225 \times 1.414 = 0.318 \, \text{nm} \] Thus, \[ \frac{a_A}{a_B} = \frac{0.390}{0.318} \approx 1.23 \] Rounded to two decimal places: \[ \boxed{\text{Lattice parameter ratio = 1.23}} \]