ii) The 100 reflection

The formula for d for cubic crystals is:

dhkl = a/Ö(h2 + k2 + l2 )

For h,k,l = 1,0,0 we have

d111 = 5.638/Ö(12 + 02 + 02 ) = 5.638/Ö1 = 5.638

From Bragg's law

l = 2d sinq

so

sin q/l = 1/2d = 1/(2 ´ 5.638) = 0.089 Å-1

The atomic scattering factors for Na⁺ and Cl^- ions can be obtained from the extract of the International Crystallographic Tables provided in the notes. The table below shows my calculated values for sin q /l = 0.089 Å^-1 (I have extrapolated from the provided values of 0.15 and 0.20). However, it turns out that it doesn't matter what the value of sin q/l is since the cosine and sine terms all cancel to zero.

Now

F(S)  =

å n

fn cos {360 (hx + ky + lz)} + i

å n

fn sin {360 (hx + ky + lz)} (in degrees),

Working out the cosine part of F₂₀₀

å n	fn cos{360 (hx + ky + lz)}	=		9.82 cos {360 (1´0 + 0´0 + 0´0)}
			+	9.82 cos {360 (1´0 + 0´1/2 + 0´1/2 )}
			+	9.82 cos {360 (1´1/2 + 0´0 + 0´1/2 )}
			+	9.82 cos {360 (1´1/2 + 0´1/2 + 0´0)}
			+	16.46 cos {360 (1´1/2 + 0´0 + 0´0)}
			+	16.46 cos {360 (1´0 + 0´1/2 + 0´0)}
			+	16.46 cos {360 (1´0 + 0´0 + 0´1/2 )}
			+	16.46 cos {360 (1´1/2 + 0´1/2 + 0´1/2 )}
		=		9.82 cos 0 + 9.82 cos 0 + 9.82 cos 180 + 9.82 cos 180
			+	16.46 cos 180 + 16.46 cos 0 + 16.46 cos 0 + 16.46 cos 180
		=		9.82 + 9.82 -9.82 - 9.82 - 16.46 + 16.46 + 16.46 -16.46
		=	0

Working out the sine part of F₂₀₀

å n	fn sin{360 (hx + ky + lz)}	=		9.82 sin {360 (1´0 + 0´0 + 0´0)}
			+	9.82 sin {360 (1´0 + 0´1/2 + 0´1/2 )}
			+	9.82 sin {360 (1´1/2 + 0´0 + 0´1/2 )}
			+	9.82 sin {360 (1´1/2 + 0´1/2 + 0´0)}
			+	16.46 sin {360 (1´1/2 + 0´0 + 0´0)}
			+	16.46 sin {360 (1´0 + 0´1/2 + 0´0)}
			+	16.46 sin {360 (1´0 + 0´0 + 0´1/2 )}
			+	16.46 sin {360 (1´1/2 + 0´1/2 + 0´1/2 )}
		=		9.82 sin 0 + 9.82 sin 0 + 9.82 sin 180 + 9.82 sin 180
			+	16.46 sin 180 + 16.46 sin 0 + 16.46 sin 0 + 16.46 sin 180
		=		0

Combining the cosine and sine parts of F₁₁₁

Fhkl2 = {cosine part}2 + {sine part}2

F1002 = 02 + 02 = 0 (Intensity)

The structure factor is zero. This means that there is no reflection by the 100 plane. By implication, the net scattering from all the electrons that comprise this 'imaginary' plane is zero. This will be due to the geometry of the crystal, inparticular brought about by symmetry elements in the crystal. Centred lattices, glide planes and screw affect the number of observable peaks and produce systematic absences that occur for specific values of hkl (meeting non-reflection conditions). The sodium chloride unit cell is cubic and as a consequence has high symmetry. This increases the likelyhood of systematic absences.

Comment: Correct: Note, in particular, that one can easily perceive in your answer the cancelling out of scattering from pairs of Na and Cl.
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