Past exam of the materials science course of the University of Cambridge / 2017 / Paper 1 / 3 / b

For a cubic crystal, $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$ and first-order bragg diffraction requires $2d_{hkl}\sin \theta =\lambda$. The face-centred lattice sum isIt is four if $h,k,l$ are all even or all odd, and zero for mixed parity. With the NaCl basis displaced by $(\frac{1}{2},0,0)$, the allowed amplitudes are $F_{hkl}=4[f_{\mathrm{Cl}}+(-1{)}^{h+k+l}{f}_{\mathrm{Na}}]$, up to the arbitrary choice of origin/species. Both allowed parity classes survive because the two ionic scattering factors are not identical. Intensities depend on $|F{|}^2$, geometry, absorption and scattering vector.

There are three separate tests: nonzero structure factor, accessible Bragg angle, and accessible plane orientation. In a symmetric reflection scan the scattering vector is normal to the specimen face, selecting the surface-parallel family and its allowed orders. Other families need a different mounting, independent orientation control or transmission/another facet. The original diagrams distinguish these geometric requirements; angles in the reciprocal-space panel are between plane normals, not Bragg angles.

The even indices give an allowed reflection. Its spacing is $d_{200}=0.280\mathrm{nm}$, andThe $(200)$ planes are parallel to the cut $(100)$ face, so the ordinary coupled reflection scan reaches them directly. This is an easy, strong peak, with the two ionic amplitudes adding rather than subtracting.

The $(111)$ reflection is allowed, with $d_{111}=0.3233\mathrm{nm}$ and $2{\theta }_B=27.56^{\circ }$. Its odd-parity amplitude is proportional to $f_{\mathrm{Cl}}-f_{\mathrm{Na}}$, generally weaker than the corresponding sum. Its normal makes ${\cos }^{-1}(1/\sqrt{3})=54.74^{\circ }$ with the $[100]$ surface normal.

It is therefore not obtained in the ordinary symmetric scan of that face, despite its modest Bragg angle. In a usual edge-aligned cubic mounting its normal also has an out-of-horizontal-plane component. A suitable multi-axis reorientation, or a crystal cut with a $(111)$ face, makes the reflection accessible; a thin specimen in transmission or a powder experiment is another route. Tilting the scattering vector does not by itself ensure front-face reflection: because $54.74^{\circ }$ exceeds ${\theta }_B=13.78^{\circ }$, incident and outgoing rays for that highly asymmetric orientation cannot both lie outside the same flat $(100)$ face. A suitable specimen shape/transmission path or recut face is needed.

The indices have mixed parity, so the face-centred lattice sum vanishes: the ideal NaCl $(300)$ reflection is systematically absent. A smaller wavelength, a larger detector range or an extra rotation axis does not remove that extinction. A nominal peak there could reflect another wavelength/reflection, multiple diffraction, another phase or broken ideal translational symmetry; it would not be an ordinary kinematic $(300)$ reflection of ideal NaCl. The allowed $(600)$ peak is a different reciprocal-lattice vector, not a recovered $(300)$ peak.

The all-even $(220)$ reflection is allowed. Here $d_{220}=0.1980\mathrm{nm}$ and $2{\theta }_B=45.77^{\circ }$. Its normal is $45^{\circ }$ from $[100]$ and can lie in the horizontal scattering plane in an edge-aligned mounting. Nevertheless it is not parallel to the large $(100)$ face, so the ordinary coupled symmetric scan misses it.

Independent sample rotation can bring the $(220)$ normal onto the Bragg bisector, but there is an important specimen constraint: ${\theta }_B=22.89^{\circ }$ is less than the $45^{\circ }$ surface inclination. One ray then traverses the slab rather than emerging from the incident face. Thus a thin transmission specimen or access through another face can permit detection after suitable remounting, whereas a thick absorbing slab in front-face reflection is unfavourable. A $(110)$-cut crystal allows a straightforward symmetric $(220)$ measurement, and a powder specimen supplies the needed orientations. Unlike $(300)$, no extinction prohibits it; whether the original portable mounting can reach it depends on independent rotation, holder clearance and the transmitting path.

The all-even indices are allowed, but $d_{640}=0.07766\mathrm{nm}$ is small. copper $K_{\alpha }$ givesThis is near backscattering and outside an ordinary portable instrument whose arm reaches only about $130^{\circ }$. The numerical angular limit should be checked on the actual model; the mathematical requirement is that its maximum $2\theta$ reach $165^{\circ }$. It also requires a normal inclined ${\tan }^{-1}(4/6)=33.69^{\circ }$ to $[100]$, so the coupled surface-normal scan is insufficient. atomic scattering factors are smaller at this larger scattering vector.

A suitable high-angle, independently orientable diffractometer could detect it. A shorter selected wavelength can instead reduce the angular requirement: with $\lambda =0.071\mathrm{nm}$, $2{\theta }_B\simeq 54.4^{\circ }$, while orientation still needs to be set appropriately. If $\lambda$ exceeded $2d=0.1553\mathrm{nm}$ there would be no first-order Bragg solution at all. The tube continuum and other characteristic lines must therefore be distinguished from the specifically calculated copper $K_{\alpha }$ peak.