Respuesta :
Answer:
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.
