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The Bravais lattice is a face-centered cubic lattice. Four inequivalent representatives of its lattice points in the conventional cubic unit cell, in fractional crystal coordinates, areBoundary copies in the drawing do not constitute additional independent lattice points. Attach the two-atom atomic basisto each lattice point. The resulting aluminium phosphide crystal structure is the zinc blende structure. For example, the four phosphorus positions reduced into the conventional unit cell are $(1/4,1/4,1/4)$, $(1/4,3/4,3/4)$, $(3/4,1/4,3/4)$, and $(3/4,3/4,1/4)$, reproducing the heights in the source drawing. There are four aluminium and four phosphorus chemical atoms per conventional unit cell, but only two chemical atoms per primitive unit cell.

Given independent primitive lattice translation vectors ${\mathbf{a}}_1,{\mathbf{a}}_2,{\mathbf{a}}_3$, defineThen ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$. The reciprocal lattice consists ofequivalently all wavevectors for which $e^{i\mathbf{G}\cdot \mathbf{R}}=1$ for every direct lattice vector $\mathbf{R}$.

For the face-centered cubic lattice, chooseTheir scalar triple product is $a^3/4$, and direct evaluation of the cross products givesEvery integer combination has coordinates $(2\pi /a)(H,K,L)$ with $H,K,L$ all even or all odd. Conversely, for such coordinates the coefficients $(K+L)/2,(H+L)/2,(H+K)/2$ are integers, so every such point occurs. The even-coordinate points are the corners of cubes of edge $4\pi /a$, and the odd-coordinate points are their body centres. ThereforeThis is a property of the Bravais lattice; the two-atom atomic basis changes structure factors rather than these reciprocal lattice translation vectors.

Use the long-wavelength linear dispersion relation $\omega \simeq v_L|\mathbf{k}|$ as an estimate. The phonon energy is thenThis extrapolates the measured low-wavevector speed of sound to the $X$ point, the first Brillouin zone boundary along $[100]$. It is not an exact phonon dispersion relation there: the discrete crystal structure generally bends the branch, and the initial slope alone cannot determine its boundary energy. In particular, $(2\pi /a,0,0)$ is not a reciprocal lattice vector of the face-centered cubic lattice; the smallest such axial vector is $(4\pi /a,0,0)$. There is therefore no reason for this acoustic phonon energy to vanish.

The ordered crystal structure consists of a primitive cubic lattice with the four-site atomic basisin fractional crystal coordinates. The drawing shows all boundary representatives of these sites. Eight corner gold chemical atoms, shared among eight unit cells each, contribute one gold chemical atom. Six face-centred copper chemical atoms, shared between two unit cells each, contribute three copper chemical atoms. Thuswhich gives the required stoichiometry. Although the unlabelled positions form a face-centered cubic lattice, a face-centering lattice translation exchanges gold and copper and therefore fails to preserve the ordered crystal structure. Its Bravais lattice is primitive cubic lattice.

Replace every coloured site in the drawing by an identical site carrying $\bar{f}$. The face-centering lattice translations now preserve the average crystal structure, whose Bravais lattice is a face-centered cubic lattice. One choice of primitive lattice translation vectors isA coherent Bragg diffraction peak requires $\mathbf{G}\cdot {\mathbf{a}}_j\in 2\pi \mathbb{Z}$ for all three primitive lattice translation vectors. Thus $k+l$, $h+l$, and $h+k$ must all be even, equivalent to all three Miller indices having the same integer parity. This is the face-centred cubic selection rule: $(110)$ is forbidden and $(111)$ is allowed. The restoration of these lattice translations explains the disappearance of the ordered superlattice reflections.

For type I, a magnetic unit cell uses the magnetic Bravais translations $(0,a,a)$, $(a,0,a)$ and $(a,a,0)$. They form an FCC lattice with conventional cubic parameter $2a$. A two-spin basis is an up spin at $(0,0,0)$ and a down spin at $(a,0,0)$, or fractional conventional coordinates $(0,0,0)$ and $(1/2,0,0)$. Every translation preserves the parity of $n_x+n_y+n_z$.

At $\mathbf{Q}=(2\pi /a)(h,k,l)$, the FCC translation sum allows $h,k,l$ all integers or all half-odd-integers. The spin basis contributes $S[1-e^{2\pi ih}]$, which removes the integer set and leavesEquivalently the spin field is $S(-1{)}^{n_x+n_y+n_z}$, so magnetic Bragg peaks are nuclear reciprocal vectors shifted by $(1/2,1/2,1/2)$.

For type II with parallel spins along $z$, use $(a,a,0)$, $(-a,a,0)$ and $(0,0,a)$ as magnetic translations: a primitive tetragonal cell with side lengths $\sqrt{2}a,\sqrt{2}a,a$, containing an up spin at zero and a down spin at $(a,0,0)$. The reciprocal translations have $h,k$ both integers or both half-odd-integers, with $l$ integer. The same basis factor removes the integer-$h$ set:This also follows from $S(-1{)}^{n_x+n_y}$. Other domains permute the three coordinate directions. A magnetic neutron cross section further multiplies these lattice conditions by the magnetic form factor and by the component of ordered moment perpendicular to $\mathbf{Q}$; that polarization factor can suppress an otherwise allowed peak.

In the conventional cubic unit cell, the four face-centered cubic lattice positions contributeThe two-atom atomic basis of a crystal gives the zinc blende structure factorMixed-parity Miller indices have $S_{\mathrm{FCC}}=0$. With all indices even or all odd, $S_{\mathrm{FCC}}=4$. For real atomic scattering factors,Hence all same-parity reflections are generally permitted in GaAs, although the last class is weak when the two atomic scattering factors nearly coincide. These are the structural selection rules; a particular experiment must also satisfy the elastic Bragg scattering condition.

Using the supplied iron parameters givesAn internal Weiss molecular field depends on temperature through $M(T)$. Its low-temperature saturation estimate uses $M_s=ng{\mu }_{B}J\simeq 1.58\times 10^6\mathrm{A}{\mathrm{m}}^{-1}$:At and above $T_C$ the zero-applied-field magnetization vanishes in this model, so this is a saturation estimate, not the field at $T_C$.

Use the paper's stipulated face-centered cubic lattice, which has twelve nearest neighbours. The saturated one-ion alignment scale isDividing the effective field among twelve equivalent neighbours gives the per-bond exchange interaction scaleTo specify the convention, write the Heisenberg model as $\mathcal{H}=-\mathcal{J}{\sum }_{⟨ij⟩}{\mathbf{J}}_i\cdot {\mathbf{J}}_j$, with dimensionless angular-momentum operators and each bond counted once. Then $g{\mu }_B{B}_e=12\mathcal{J}⟨J_z⟩$, so $\mathcal{J}=3k_B{T}_C/[12J(J+1)]\simeq 11.2\mathrm{meV}$ here. A convention with $-2{\mathcal{J}}^{\prime }$ instead has ${\mathcal{J}}^{\prime }=5.62\mathrm{meV}$. The total saturated interaction energy per ion is $-12\mathcal{J}{J}^2/2\simeq -67.4\mathrm{meV}$; the factor of one half avoids counting each bond twice.

The fluorite structure consists of a face-centered cubic lattice of calcium sites with all eight tetrahedral interstitial sites occupied by fluoride. In fractional coordinates of the conventional cubic unit cell, the calcium basis is $(0,0,0),(0,\frac{1}{2},\frac{1}{2}),(\frac{1}{2},0,\frac{1}{2}),(\frac{1}{2},\frac{1}{2},0)$, and fluoride occupies all combinations $(x,y,z)$ with $x,y,z\in \{\frac{1}{4},\frac{3}{4}\}$. The projection therefore superposes two fluoride sites at each quarter-coordinate position. Labels in the original diagram give fractional depth along $[001]$; periodic boundary sites are displayed on both edges.

Counting shared sites gives $8/8+6/2=4$ calcium sites and eight interior fluoride sites: four formula units per conventional cell.

The nickel-rich $\gamma$ matrix is a disordered substitutional solid solution on a face-centered cubic lattice. The ordered ${\gamma }^{\prime }$ phase has the L12 structure, approximately $\mathrm{N}{\mathrm{i}}_3\mathrm{Al}$: aluminium occupies cube corners and nickel occupies face centres. Its occupied-site geometry is often called ordered FCC, but its chemically labelled Bravais lattice is primitive cubic, with a four-atom basis. A face-centre translation exchanges chemical sublattices and is not a symmetry of the ordered crystal.

Each Al atom has twelve nearest-neighbour Ni atoms at Euclidean distance $a/\sqrt{2}$. Each Ni atom has twelve nearest neighbours too, but these are four Al and eight Ni. The chemical distinction matters for slip. In disordered $\gamma$, the shortest perfect Burgers vector isFor the ordered primitive cubic translation lattice, the shortest order-preserving perfect vector isThis does not imply that $⟨100⟩$ slip must dominate. A vector must lie in a favourable glide plane and have an accessible core structure and activation stress. Matrix dislocations commonly glide on $\{111\}⟨110⟩$ systems. A single $a⟨110⟩/2$ passage through ${\gamma }^{\prime }$ leaves an antiphase boundary; a second matching passage restores order, giving total vector $a⟨110⟩$. Such a superdislocation has a longer total vector than $a⟨100⟩$, yet its constituent dislocations can move on the favourable close-packed planes and interact with the matrix. Paired motion balances line and antiphase-boundary energies, so the shortest abstract lattice translation alone does not select the active slip system.^[ref]

For a disordered face-centered cubic lattice, equal average scattering factors on the four sites give reflections only when $h,k,l$ are all even or all odd. Bragg's law also requires $h^2+k^2+l^2\le 4a^2/{\lambda }^2$. With Fe K$\alpha$ wavelength $\lambda \simeq 0.1936\mathrm{nm}$, the upper bound is $15.41$. The allowed sums are $3,4,8,11,12$, givingThe next allowed family, $400$, has squared-index sum $16$ and cannot satisfy the diffraction condition for this wavelength. This list assumes the full geometrically accessible angular range, rather than an instrument-specific cutoff.

The fluorite structure has Ce on the corner and face-centre sites of a face-centered cubic lattice. Oxygen occupies all eight positions $(x,y,z)$ with each coordinate independently $1/4$ or $3/4$. In projection, there are four oxygen positions $(1/4,1/4),(1/4,3/4),(3/4,1/4),(3/4,3/4)$, each representing two heights. Accounting for shared boundary sites gives four Ce and eight O per conventional unit cell.

In the kinematic approximation, the face-centered cubic lattice contributes only reflections with $h,k,l$ all even or all odd. Viewing along $[001]$ selects the zero-order zone $l=0$, so the nonzero allowed spots within the cutoff are $(\pm 2,0,0),(0,\pm 2,0),(\pm 2,\pm 2,0)$, with the transmitted beam at the centre. The nearest-neighbour separation is $2/a_{\mathrm{Al}}$ in the crystallographic reciprocal-distance convention, or $4\pi /a_{\mathrm{Al}}$ in the angular-wavevector convention. No matrix lattice parameter is supplied, so a numerical matrix spacing would require extra data.

A Bravais lattice is the set of all integer combinations of linearly independent primitive translation vectors. Its reciprocal lattice consists of $g$ satisfying $g\cdot l\in 2\pi \mathbb{Z}$ for every translation $l$; equivalently $e^{ig\cdot l}=1$.

The original PDF's third family has a bare $j$, not $n_{2}j$. Taken literally, the listed set is not the claimed Bravais lattice: the claimed primitive vector $a(i+k)/2$ does not occur. Its half-integer first and third coordinates force it into that third family, whose second coordinate is always $a$, not zero. The intended repair is $j\mapsto n_{2}j$ in the third family. The following conclusions apply to that corrected set.

The four corrected families are exactlyThe allowed parity patterns are all even, or exactly two odd coordinates. An integer combination of the proposed primitive vectors has coordinates $(a/2)(m_2+m_3,m_1+m_3,m_1+m_2)$, so lies in this set. Conversely, solveAll three are integers precisely when $h+k+l$ is even. This proves that the corrected set is the face-centered cubic lattice with the proposed primitive basis.

Direct dot products give $a_i\cdot b_j=2\pi {\delta }_{ij}$ forThese vectors generate the entire reciprocal lattice, since the coefficients of an arbitrary reciprocal vector in this basis are its integer pairings with the $a_i$. Its vectors are $(2\pi /a)(h,k,l)$ with all three coordinates of the same parity, the body-centered cubic lattice.

In Bragg scattering, translating a scattering atom by $l$ changes the amplitude by $e^{i(k-k^{\prime })\cdot l}$. Constructive interference from every lattice translate requires this factor to be one, so $k^{\prime }-k=g$ is a reciprocal lattice vector. For elastic scattering, $|k^{\prime }|=|k|$; the chord between these two vectors has length $2|k|\sin (\theta /2)$. ThereforeThe shortest nonzero reciprocal vectors are the $(111)$ family of length $(2\pi /a)\sqrt{3}$, followed by the $(200)$ family of length $(2\pi /a)2$. Identical atoms on this Bravais lattice introduce no additional basis extinction. If both families are accessible at the incident wavelength, their scattering angles obey $\boxed{\sin ({\theta }_1/2)/\sin ({\theta }_2/2)=\sqrt{3}/2}$.

The reciprocal lattice is ${\Lambda }^{\ast }=\{G:G\cdot \ell \in 2\pi \mathbb{Z}\text{ for every }\ell \in \Lambda \}$. With oriented primitive-cell volume $v_c=a_1\cdot (a_2\times a_3)$, its basis is $b_1=2\pi (a_2\times a_3)/v_c$ and the cyclic variants; thus $b_i\cdot a_j=2\pi {\delta }_{ij}$.

The total potential is $V(x)={\sum }_{\ell }U(x-\ell )$. Using the Fourier transform convention $\tilde{U}(q)=\int e^{-iq\cdot x}U(x)d^{3}x$, translation gives $\tilde{V}(q)=\tilde{U}(q)\Delta (q)$, withThe given Born approximation therefore supplies the amplitude $\boxed{-m\Delta (q)\tilde{U}(q)/(2\pi {\hslash }^2)}$. A finite geometric series gives, when the denominators are nonzero,At reciprocal-lattice vectors each phase is one, so $|\Delta |=N_1{N}_2{N}_3$, the number of atoms. For large finite crystals these narrow peaks represent coherent Bragg scattering, with peak intensity proportional to the squared number of atoms; their finite width reflects the finite crystal size.

For the specified face-centered cubic lattice, $v_c=a^3/4$ andThe reciprocal lattice have coordinates $(2\pi /a)(h,k,l)$ with $h,k,l$ all even or all odd. The two shortest nonzero length families are $G_1=2\pi \sqrt{3}/a$ for $(111)$ and $G_2=4\pi /a$ for $(200)$. Elastic scattering satisfies $|q|=2|k|\sin (\theta /2)$. The assumed incident magnitude exceeds $G_2/2$, so both families are kinematically accessible, andThis is the angular prediction for suitable crystal orientations, or for a powder containing those orientations. A fixed single-crystal orientation must additionally satisfy the vector Elastic Bragg scattering condition $2k\cdot G+|G{|}^2=0$; the magnitude bound alone cannot guarantee both peaks for every incident direction.

Symmetry breaking in nanocrystal growth makes growth rates along initially equivalent crystallographic directions unequal, so the final shape has lower symmetry than an unperturbed seed. It need not change the underlying face-centered cubic lattice. A cubic lattice does not require a cubic external shape.

Anisotropy can originate in a twin or stacking fault, unequal ligand coverage of nominally equivalent sites, selective adsorption or etching, a substrate or external field, or a local fluctuation amplified because newly deposited material changes subsequent growth. Facet-selective binding suppresses some normal growth rates; concentration and reduction kinetics determine whether atoms can redistribute towards a near-equilibrium shape or remain in a kinetically selected one. A facet with slow normal growth tends to remain exposed, while a fast-growing facet can disappear.

For a one-dimensional silver example, begin with a small seed with multiple regions related by crystal twinning containing five sectors around a common axis. Ligand-protected lateral faces and more active ends favour axial addition, producing an elongated pentagonal rod and then a wire. The defect geometry and surface kinetics jointly distinguish the axial direction. Fivefold twinning is a composite structure with strain accommodation, not a perfect single fcc crystal with forbidden fivefold lattice symmetry.

For a two-dimensional example, begin with a seed containing a planar twin or stacking fault. Basal-face protection and enhanced incorporation at suitable edges favour lateral extension over thickening, producing a thin triangular or hexagonal plate. The accompanying schematic emphasizes seed, symmetry-breaking event and anisotropic growth stages. These are representative routes, not universal requirements for every rod or plate. Experimentally, silver-seed growth can also be redirected into lower-symmetry forms by controlling reduction kinetics.^[ref]

The characteristic effect is an irradiation-induced upward shift of the ductile-to-brittle transition in ferritic steel. A component previously used above its transition can then fail by cleavage fracture at the same service temperature, with much less warning from plastic deformation.

Ferrite has a body-centered cubic lattice. At low temperature, dislocation glide becomes difficult enough that brittle crack propagation can precede substantial plastic relaxation. Irradiation hardening raises the yield strength by introducing obstacles to dislocation glide; the competition with cleavage fracture consequently shifts toward brittle failure. Austenite, with its face-centered cubic lattice, does not normally show this same sharp cleavage transition. This does not make austenitic steel immune to loss of ductility, helium embrittlement or other radiation damage in materials.

Examine the fracture surface for cleavage fracture facets and river patterns, and compare Charpy impact test energy-versus-temperature curves from exposed material and an unirradiated reference of the same composition and heat treatment. The reference is important: a low impact energy from one failed specimen cannot establish an irradiation-induced shift. Transition curves can also use lateral expansion or the proportion of ductile fracture appearance.^[ref]

A direct fracture toughness test over the transition region complements the Charpy impact test; the two measurements are not interchangeable. Combine these observations with exposure history, local temperature, microstructure and crack location. Brittle appearance supports the failure mechanism, while a measured transition shift and suitable controls support its attribution to irradiation.

Assume a homogeneous Fe–Cr–Ni alloy, normal incidence, independent scattering events and the stated effective scattering cross-sections. Neglect absorption, other constituents and changes in neutron energy within the thin sheet. The atomic fractions are not supplied; take $x_{\mathrm{Fe}}=0.70$, $x_{\mathrm{Cr}}=0.20$, $x_{\mathrm{Ni}}=0.10$ as an explicit representative assumption, not a composition measured or specified in the paper.

The conventional cell of a face-centered cubic lattice containsatoms. The number density is thereforeFor a mixture, the scattering probability per unit path length is ${\Sigma }_s={\sum }_j{n}_j{\sigma }_j=n{\bar{\sigma }}_s$, whereThus ${\Sigma }_s=27.751{\mathrm{m}}^{-1}$. For thickness $t=10^{-3}\mathrm{m}$, the unscattered fraction $U$ satisfies $dU/dz=-{\Sigma }_{s}U$ with $U(0)=1$. The fraction undergoing at least one neutron scattering event isThe thin-sheet approximation gives $P_s\simeq {\Sigma }_{s}t=2.78\%$, consistent to the accuracy of the assumed composition. A simple iron-dominated estimate ${\bar{\sigma }}_s\simeq 3.2$ barns gives essentially the same about $3\%$ scattered conclusion.

For any other assumed composition, use the displayed weighted formula; mass fractions must first be converted to atomic fractions. Since the supplied elemental cross-sections span $2.8$–$4.1$ barns, any mixture of these three elements at the stated number density gives approximately $2.39$–$3.48\%$. This is a bound within the idealized mixture model, not a claim that every such composition is stable austenite.

For hard spheres of radius r in a face-centered cubic lattice, contact occurs along a face diagonal: $4r=\sqrt{2}a$. Four spheres belong to a conventional cell, so the hard-sphere cubic packing fractions giveFor a body-centered cubic lattice, contact occurs along a body diagonal, $4r=\sqrt{3}a$, and there are two spheres per conventional cell:Their ratio $8/(3\sqrt{6})=1.0887>1$ proves that FCC packs more densely. Coordination alone would not establish the occupied volume fraction.

For conventional cubic cell side $a$, the face-centered cubic lattice has corner and face-centre sites, giving four lattice points per cell; the body-centered cubic lattice has corners and a body centre, giving two. Their primitive unit cells each contain one lattice point. With reciprocal vectors written $\mathbf{G}=(2\pi /a)(h,k,l)$, face-centered cubic lattice allows $h,k,l$ all even or all odd, while body-centred cubic allows $h+k+l$ even. HenceHere $a^{\ast }$ is the side of the reciprocal lattice's conventional cubic cell. The parity conditions describe the reciprocal point sets; reciprocal space contains wavevectors, not a second arrangement of physical atoms.