Structure Factor Calculation for 200 Reflection of NaCl

Structure Factor Calculation for 200 Reflection of NaCl

Using Crystallography and Bragg's law we can calculate the d spacing and other parameters for 200 reflection:

d_hkl = a/√(h² + k² + l² ) For hkl = 200 we have d₂₀₀ = 5.638/√(2² + 0² + 0² ) = 5.638/√4 = 2.819 Å From Bragg's law: λ = 2d sinθ so sinθ / λ = 1 / 2d = 1 / (2 × 2.819) = 0.177 Å^-1

The next step is to work out the atomic scattering factors (f_n) that will be needed. The course material already provided values for the atomic scattering factors for Na⁺ and Cl^- ions (obtained from the International Tables for Crystallography) in the vicinity (0.15 and 0.20) of sinθ / λ = 0.177. So interpolating between these values you can obtain the required values given in bold in the middle column of the following table:

sinθ / λ	0.15	0.177	0.20
f_Na+	9.035	8.69	8.39
f_Cl−	14.12	13.08	12.20

Note that the bold values given above were obtained by interpolating between the side-values at a ratio of 23:27 and are therefore slightly closer to the right-hand side values, but you will probably have noticed that 0.177 is almost exactly the mean (0.175) of 0.15 and 0.20 and therefore a simple mean, (9.035 + 8.39)/2 = 8.71 and (14.12 + 12.20)/2 = 13.16, would have been good enough.

We now put these atomic scattering factors into the structure factor equation given previously:

F(S) =

Σ
n

f_n exp{2π i (hx + ky + lz)}

It is easier to deal with this equation in two parts, first taking the cosine part which has 8 terms corresponding to the 8 atoms involved. Inserting the appropriate values (hkl = 200; and coordinates) and expressing in radians leads to:

Σ n	f_n cos{2π (hx + ky + lz)}	=		8.69 cos {2π (2 × 0 + 0 × 0 + 0 × 0)}
			+	8.69 cos {2π (2 × 0 + 0 × 1/2 + 0 × 1/2)}
			+	8.69 cos {2π (2 × 1/2 + 0 × 0 + 0 × 1/2 )}
			+	8.69 cos {2π (2 × 1/2 + 0 × 1/2 + 0 × 0)}
			+	13.08 cos {2π (2 × 1/2 + 0 × 0 + 0 × 0)}
			+	13.08 cos {2π (2 × 0 + 0 × 1/2 + 0 × 0)}
			+	13.08 cos {2π (2 × 0 + 0 × 0 + 0 × 1/2)}
			+	13.08 cos {2π (2 × 1/2 + 0 × 1/2 + 0 × 1/2)}
		=		8.69 cos 0 + 8.69 cos 0 + 8.69 cos 2π + 8.69 cos 2π
			+	13.08 cos 2π + 13.08 cos 0 + 13.08 cos 0 + 13.08 cos 2π
		=		8.69 + 8.69 + 8.69 + 8.69 + 13.08 + 13.08 + 13.08 + 13.08
		=	+	87.08

Similarly, working out the sine part of F₂₀₀ gives:

Σ n	f_n sin {2π (hx + ky + lz)}	=		8.69 sin {2π (2 × 0 + 0 × 0 + 0 × 0)}
			+	8.69 sin {2π (2 × 0 + 0 × 1/2 + 0 × 1/2)}
			+	8.69 sin {2π (2 × 1/2 + 0 × 0 + 0 × 1/2)}
			+	8.69 sin {2π (2 × 1/2 + 0 × 1/2 + 0 × 0)}
			+	13.08 sin {2π (2 × 1/2 + 1 × 0 + 1 × 0)}
			+	13.08 sin {2π (2 × 0 + 0 × 1/2 + 0 × 0)}
			+	13.08 sin {2π (2 × 0 + 0 × 0 + 0 × 1/2)}
			+	13.08 sin {2π (2 × 1/2 + 0 × 1/2 + 0 × 1/2)}
		=		8.69 sin 0 + 8.69 sin 0 + 8.69 sin 2π + 8.69 sin 2π
			+	13.08 sin 2π + 13.08 sin 0 + 13.08 sin 0 + 13.08 sin 2π
		=		0 + 0 + 0 + 0 + 0 + 0 + 0 + 0
		=		0

Combining the cosine and sine parts of F₂₀₀ into the complex number form gives:

F₂₀₀ = +87.08 + 0i which on squaring for the intensity leads to:

I₂₀₀ = F₂₀₀² = 7583 units

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Author(s): Paul Barnes