BCC iron has a lattice constant of 0.2856nm. Compute: a. Atomic radius of iron. b. Using a first-order reflection x-ray, wavelength 0.1542 nm, compute the expected diffraction angle (2θ) for the (200) plane. c. Will the (200) plane diffraction peak be visible in the XRD measurement?

a) Atomic radius of iron, R = 0.124 nm

b) Angle of diffraction, 2θ = 65.36°

c) The (200) plane diffraction will be visible in the XRD measurement because the angle of diffraction falls within the measurable range for XRD analysis.

Explanation:

Relationship between lattice parameter, a and atomic radius, R, for BCC structure is given by

a = 4R/√3

R = (a√3)/4

a = 0.2856nm = 2.856 × 10⁻¹⁰ m

R = (2.856 × 10⁻¹⁰)(√3)/4 = 1.24 × 10⁻¹⁰ m = 0.124 nm

b) To obtain this angle of diffraction for plane (hkl), we need some formulas

√(h² + k² + l²) = a/dₕₖₗ

dₕₖₗ = interplanar spacing, a = lattice parameter

dₕₖₗ = a/√(h² + k² + l²) = (2.856 × 10⁻¹⁰)/√(2² + 0² + 0²) = 1.428 × 10⁻¹⁰ m = 0.1428 nm

Then,

dₕₖₗ = nλ/2 sin θ

where θ = half of the diffraction angle, n is the order of reflection = 1 and λ is the wavelength of the monochromatic radiation = 0.1542 nm.

0.1428 = 1×0.1542/(2sinθ)

Sin θ = 0.5399

θ = sin⁻¹ (0.5399) = 32.68°

Angle of diffraction = 2 × 32.68° = 65.36°

c) For the plane's diffraction to be measurable, the angle of diffraction should lie between ~5° to 70°,

And 65.36° conveniently lies within this range, so, the (200) plane diffraction will be visible in the XRD measurement.

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