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The two sites give phases one and $\exp [i\pi (h+k+l)]=(-1{)}^{h+k+l}$. ThusFor real nonresonant form factors, even $h+k+l$ gives the square of the sum and odd $h+k+l$ the square of the difference. The factor $K_{hkl}$ represents the remaining experimental corrections, not another basis-interference term. Since the two ions differ, a half-body-diagonal translation does not preserve the decorated crystal. Its Bravais lattice is therefore simple cubic, with a two-ion basis; CsCl is not a monatomic body-centered cubic Bravais lattice.

Replacing both sites by identical Fe atoms changes the amplitude toNow the body-centering translation does preserve the structure. It is a body-centered cubic lattice with one Fe atom per primitive cell, or two per conventional cubic cell.

For 111 the parity is odd: CsCl generally has a nonzero amplitude $f_{\mathrm{C}{\mathrm{s}}^{+}}-f_{\mathrm{C}{\mathrm{l}}^{-}}$, whereas the identical-Fe 111 reflection is systematically absent. A numerical absolute CsCl intensity requires the angle-dependent form factors and experimental normalization; no such values are supplied. Forward electron counts, 54 for Cs+ and 18 for Cl-, explain why their amplitudes are substantially unequal, but are not substituted for the actual finite-angle factors without an additional approximation.

Separate the entire-system atom index into a cell translation $\mathbf{R}$ and an index $p$ within a single primitive unit cell. The potential becomesThe same $u_p$ is used in every translated cell because the atomic basis repeats. Normalize plane waves in volume $\Omega$ by $⟨\mathbf{x}|\mathbf{k}⟩={\Omega }^{-1/2}{e}^{i\mathbf{k}\cdot \mathbf{x}}$. For $\mathbf{G}={\mathbf{k}}^{\prime }-\mathbf{k}$, substitution and $\mathbf{s}=\mathbf{x}-\mathbf{R}-{\mathbf{r}}_p$ giveFor a real static scalar potential, define the atomic scattering factor in the positive-exponent convention of the question byThe second bracket in the matrix element is $S(\mathbf{G}{)}^{\ast }$. At a reciprocal lattice vector every term of the first bracket is one, giving the number of cells $N_c$. Therefore the stated transition-rate rule, equivalently the first Born approximation and Fermi's golden rule up to the final-state kinematic factors, yieldsThe harmless complex conjugation records the distinction between the source's positive-exponent Fourier transform and the usual positive-exponent incoming plane wave. Reversing every Fourier sign gives the equivalent amplitude convention. Only the cell-translation sum ranges over the full crystal; the crystal basis structure factor contains the atoms of one cell.

The leading coefficient fixes the atom count:With molar mass $40\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, the sample mass isOne atom belongs to each primitive unit cell of monatomic argon on a face-centered cubic lattice, but the conventional cube contains four atoms. If its side is $a$, $N=4V/a^3$, and nearest neighbors lie across half a face diagonal. Thereforeusing $V=21\times 10^{-6}{\mathrm{m}}^3$. The corresponding conventional lattice constant is $a\simeq 5.18\text{A˚}$. Treating the source's one-atom cell as the conventional cube would omit the factor four and give a wrong nearest-neighbor distance.

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.

Subtracting the three acoustic phonon branches from the six displacement normal modes leavesAt small wavevector, these involve relative motion of the two atomic basis sites rather than a rigid translation of the primitive unit cell. Their angular frequencies remain nonzero at the zone centre. The count uses the primitive unit cell; using all eight chemical atoms in the conventional unit cell would spuriously multiply the number of distinct branches in the first Brillouin zone.

There are two chemical atoms in the primitive unit cell, hence six displacement coordinates and six phonon branches at each wavevector. Three rigid translations cost no restoring potential energy in the long-wavelength limit. ConsequentlyFor propagation along $[100]$, one acoustic phonon has longitudinal polarization along $x$ and two have transverse polarization along $y,z$. The cubic crystal symmetry makes those two transverse directions equivalent. In terms of the elastic stiffness tensor and the mass density,Thus there are two distinct acoustic velocities in general, one longitudinal and a doubly degenerate transverse value. This degeneracy of normal modes concerns the two transverse polarizations; it does not imply equality with the longitudinal speed of sound.

Without the Peierls distortion, adjacent chemical atoms are separated by $a/2$. The primitive unit cell contains one chemical atom and contributes one electron. A energy band with two spin states accommodates two electrons per primitive unit cell, so the first energy band is half filled. Its true first Brillouin zone extends to $\pm 2\pi /a$, while the Fermi points are at $\pm \pi /a$.

Describing the same chain with a doubled unit cell of length $a$ merely folds the half-filled energy band into two branches that touch at those Fermi points. The unbroken lattice translation by $a/2$ prevents an artificial band gap there. A Peierls distortion breaks that smaller lattice translation and supplies the missing Fourier coefficient at $2k_F=G$; the lower folded energy band can then be completely filled and separated from the upper energy band. Thus the zero-band gap result at $\delta =0$ agrees with the original one-electron band filling.

A phonon is one quantum of a normal vibrational mode of a crystal. In the harmonic approximation, each mode with wavevector $\mathbf{q}$, branch index $s$ and angular frequency ${\omega }_s(\mathbf{q})$ is an oscillator with energies $\hslash {\omega }_s(n+1/2)$. A phonon changes the occupation $n$ by one; it is not a particular atom moving through the solid.

For a three-dimensional primitive cell containing $r$ atoms there are $3r$ branches. Three acoustic branches approach zero frequency as $\mathbf{q}\to 0$: at zero wavevector they represent rigid translations, with all basis atoms displaced together. Their small-wavevector frequencies usually grow linearly with wavevector. The remaining $3r-3$ optical branches have internal relative motion of basis atoms and generally nonzero zone-centre frequencies. The names reflect their coupling to radiation in suitable polar crystals, not a rule that every optical mode must be infrared active. Acoustic and optical branches can each have longitudinal or transverse polarizations.

The primitive unit cell contains two carbon atoms and four edge-shared oxygen atoms, each contributing one half: $n=2+4/2=4$ atoms. Equivalently, one molecular repeat contributes its two carbon atoms but only half of each of its four shared oxygen atoms. There are therefore $3n=12$ phonon branches.

At the Brillouin zone centre, three rigid translations give three acoustic phonon branches; the remaining $3n-3$ are optical phonon branches:A free layer has two in-plane acoustic modes and an out-of-plane flexural acoustic mode. A two-dimensional lattice does not imply that its atoms can move only in two directions. Artificially suppressing all out-of-plane motion would instead define a different model with two acoustic and six optical branches.

The primitive translations connect the two centred-rectangular sublattices. In the orientations marked in the source, ${\mathbf{a}}_p$ points right and down, whereas ${\mathbf{b}}_p$ points right and up. ThusTheir parallelogram has half the area of the rectangular cell, confirming that the latter contains two primitive unit cells.

For ordinary one-phonon infrared optical absorption, both an energy and a momentum selection rule must hold. The optical photon has negligible crystal momentum on the scale of the Brillouin zone, so crystal momentum conservation selects $\mathbf{q}\simeq 0$ modulo a genuine reciprocal lattice vector. In addition, the mode must change the macroscopic electric polarization, with a component along the incident polarization of an electromagnetic wave.

In terms of atomic displacements ${\mathbf{u}}_j$ and Born effective charge tensors $Z_j^{\ast }$, the relevant derivative is proportional to ${\sum }_j{Z}_j^{\ast }{\mathbf{u}}_j$ per primitive unit cell. Thus an optical phonon at $\mathbf{q}=0$ is not automatically infrared active: its dipole contributions may cancel. Disorder-assisted and multiphonon processes relax these simple first-order selection rules.

Write $\mathbf{r}={\sum }_i{s}_i{\mathbf{a}}_i$. The electron density is periodic with period one in each $s_i$, so its three-dimensional Fourier series has integer harmonics:If $\Omega$ is a primitive unit cell of volume ${\Omega }_c$, Fourier orthogonality givesConversely, translation by ${\mathbf{a}}_i$ multiplies a general plane wave by $e^{i\mathbf{q}\cdot {\mathbf{a}}_i}$. Periodicity permits only $\mathbf{q}\cdot {\mathbf{a}}_i\in 2\pi \mathbb{Z}$, precisely the reciprocal lattice vectors. For a physical bounded periodic electron density, the expansion exists in the usual square-integrable sense.

There are two electrons per primitive unit cell; a spin-degenerate energy band accommodates exactly that number. A semiconductor results when the first band is full and does not overlap the second. In the leading two-wave/four-wave approximation, compare the first-band maximum at $M$ with the second-band minimum at $X$:This is an indirect band gap; a nonzero local splitting at $X$ alone is insufficient. At the onset,With $a=0.3\mathrm{nm}$, ${\hslash }^2/(2m_e)=0.03810\mathrm{eV}\mathrm{n}{\mathrm{m}}^2$ gives $E_X\simeq 4.178\mathrm{eV}$, henceFor $V_{10}$ more negative than this value the estimated indirect band gap is positive; at equality the bands just touch. This is the onset estimate within the source's nearly-free electron model truncation, not an exact critical potential after coupling all plane waves.

A Bravais lattice is a set of equivalent points generated by integer combinations of three independent translation vectors. A crystal motif is the arrangement of atoms attached identically to every lattice point. Together they specify the periodic crystal structure: lattice translations repeat the motif without changing its internal arrangement.

A unit cell is a region whose translations tile the structure. A primitive unit cell contains one lattice point after sharing boundary points with neighboring cells. A conventional cell can contain more than one lattice point to display the crystal's symmetry with simple axes and angles; for example, a conventional face-centered cube has four lattice points but a less visually symmetric primitive cell has one.

Rotational symmetry compatible with a periodic three-dimensional lattice has order $1,2,3,4$ or $6$, as expressed by the crystallographic restriction theorem. A crystallographic fivefold axis is excluded from an ordinary periodic crystal. This statement concerns exact lattice symmetry; it does not exclude local fivefold atomic arrangements or nonperiodic ordered structures.

A reciprocal vector must have integral multiples of $2\pi$ as its dot products with lattice translations. The translations along the conventional axes require $G=2\pi (h/a,k/a,l/b)$, and the body-centering translation requires $h+k+l$ to be even. ThusThis is a face-centered tetragonal lattice, with conventional dimensions $4\pi /a,4\pi /a,4\pi /b$. The primitive unit cell volumes are $\boxed{a^{2}b/2}$ in direct space and $\boxed{2(2\pi {)}^3/(a^{2}b)}$ in reciprocal space. Their conventional cell volumes are $a^{2}b$ and $8(2\pi {)}^3/(a^{2}b)$, containing two and four lattice points respectively.

The listed plane is $x/a+z/b=1$, whose normal reciprocal vector is $\boxed{G_0=2\pi (1/a,0,1/b)}$. It is an allowed primitive vector in that direction; half of it has nonintegral indices. The plane spacing is $d=2\pi /|G_0|=(a^{-2}+b^{-2}{)}^{-1/2}=4a/5$ when $b=4a/3$. Bragg's law gives $2d\sin \theta =m\lambda$, soThese are approximately $18.21^{\circ },38.68^{\circ },69.64^{\circ }$. Here $\theta$ is the glancing Bragg angle; the angle between incident and scattered beams is $2\theta$.

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.

Put type-A atoms at $2ja$ and type-B atoms at $(2j+1)a$. Their displacements $u_j,v_j$ obey Newton's second law,Choose normal modes $u_j=U{e}^{i(2jaq-\omega t)}$ and $v_j=V{e}^{i((2j+1)aq-\omega t)}$. Substitution givesA nonzero displacement vector requires zero determinant:Solving this quadratic in ${\omega }^2$ gives the diatomic-chain dispersion relation with equal springs,The primitive unit cell has length $2a$, so the first Brillouin zone is $|q|\le \pi /(2a)$.

Take nearest-neighbour separation $d$, a common spring constant $K$, and one scalar displacement per atom. For a monatomic chain, the harmonic approximation near a potential minimum, Newton's second law and a normal mode giveThe monatomic-chain dispersion relation has one acoustic phonon branch, with $\omega \simeq d\sqrt{K/m}|k|$ near the origin. The first Brillouin zone ends at $\boxed{k=\pm \pi /d}$, where $\omega =2\sqrt{K/m}$ and the group velocity vanishes.

For alternating masses $m,M$, choose a two-atom cell of length $2d$. The coupled normal modes give the diatomic-chain dispersion relation with equal springs,The first Brillouin zone now ends at $\boxed{k=\pm \pi /(2d)}$. The minus branch is the acoustic phonon branch; its two atoms move together at long wavelength. The plus branch is the optical phonon branch; at $k=0$ the atoms move oppositely with a stationary centre of mass and ${\omega }_{+}^2=2K(1/m+1/M)$. For $M>m$ the edge frequencies are $\sqrt{2K/M}$ and $\sqrt{2K/m}$, with a gap between them.

If $M=m$, the two branches reduce to $2\sqrt{K/m}|\sin (kd/2)|$ and $2\sqrt{K/m}|\cos (kd/2)|$, touching at $\sqrt{2K/m}$. This is zone folding in an equal-mass diatomic chain: identical sites actually have a primitive unit cell of length $d$. A doubled cell folds the monatomic branch but opens no gap. Equal masses alone do not specify all possible spring models; the sketches assume equal nearest-neighbour spring constants.

A Bravais lattice is the set of positions generated by integer combinations of independent primitive translation vectors, $\mathbf{R}=n_1{\mathbf{a}}_1+n_2{\mathbf{a}}_2+n_3{\mathbf{a}}_3$. Every lattice point has the same translational environment. A crystal may additionally have a basis of several atoms attached to each lattice point; those atoms are not automatically separate points of its Bravais lattice.

A unit cell is a region whose translated copies fill the crystal without gaps or overlaps. It can be chosen for convenient symmetry, and a conventional cell may contain several lattice points. A primitive unit cell is a smallest such repeating cell, containing exactly one lattice point when boundary sharing is counted. Its shape is not unique, but its volume $\Omega =|{\mathbf{a}}_1\cdot ({\mathbf{a}}_2\times {\mathbf{a}}_3)|$ is fixed for a given lattice. For a monatomic Bravais crystal, each primitive cell contains one atom.

Interpret the sphere as an atom of iron, and assume neighbouring atoms touch without overlap. Its radius is $r=0.124\mathrm{nm}=1.24\times 10^{-10}\mathrm{m}$, and its mass is $m_{\mathrm{Fe}}=56u\simeq 9.30\times 10^{-26}\mathrm{kg}$, using atomic mass unit $u=1.66054\times 10^{-27}\mathrm{kg}$.

In the face-centered cubic lattice, atoms touch along a face diagonal: $\sqrt{2}{a}_F=4r$, so $a_F=2\sqrt{2}r=0.3507\mathrm{nm}$. Four atoms occupy a conventional cube. The hard-sphere atomic volume in cubic lattices therefore givesIn the body-centered cubic lattice, atoms touch along the body diagonal: $\sqrt{3}{a}_B=4r$, so $a_B=4r/\sqrt{3}=0.2864\mathrm{nm}$. There are two atoms per conventional cube, givingThese are primitive unit cell volumes, not the volumes of the conventional cubes or of the atomic spheres. The denser FCC contact packing gives the higher mass density in this fixed-radius model. Actual hot phases have temperature-dependent dimensions and bonding, so these numbers should not be treated as exact experimental phase densities.