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Choose a primitive cubic lattice with $a=0.400\mathrm{nm}$, one identical point scatterer per primitive unit cell, incident direction ${\mathbf{s}}_{\mathrm{in}}=(0,0,1)$, and the continuous wavelength band $0.080\le \lambda \le 0.160\mathrm{nm}$. This is a hypothetical model, not the zinc blende specimen of the Laue X-ray diffraction experiment of 1912. Write a candidate reciprocal lattice vector as $\mathbf{G}=(2\pi /a)(h,k,-m)$ with $m>0$, and set $S=h^2+k^2+m^2$. The Laue equations and elastic scattering giveA transmission photographic plate at $z=L>0$ intercepts only $s_{\mathrm{out},z}>0$. Its coordinates relative to the direct beam areFor $(h,k,-m)=(2,1,-1)$ and $L=50\mathrm{mm}$, $S=6$, $\lambda =0.13333\dots \mathrm{nm}$ and ${\mathbf{s}}_{\mathrm{out}}=(2/3,1/3,2/3)$, so $(X,Y)=(50,25)\mathrm{mm}$. Its length is one, and each component difference satisfies the corresponding Laue equations. This verifies a direction and a selected wavelength, without predicting an exposure or measured wave intensity.

The first panel has fourfold symmetry because its chosen Bravais lattice, beam axis and square detector share that symmetry. For the tilted panel, apply a rotation matrix $R$ to each index vector, set $v=R(h,k,l)$, and use $\lambda =-2a{v}_z/|v{|}^2$ and ${\mathbf{s}}_{\mathrm{out}}=(0,0,1)+(\lambda /a)v$. All indices with $|(h,k,l)|\le 2a/{\lambda }_{\min }=10$ suffice: elastic scattering bounds $|\mathbf{G}|\le 2|{\mathbf{k}}_{\mathrm{in}}|$. Thus the plotted spectral and angular selection is complete within this ideal point model and detector window. Multiples of an index vector select inversely scaled wavelengths and can land on the same spot; a white-beam spot need not identify a unique diffraction order.

The electronic bandwidth is the difference between the maximum and minimum of one energy band over the Brillouin zone. It has units of energy, unlike a frequency bandwidth in signal processing. A nearest-neighbour tight-binding model on a primitive cubic lattice with equal real hopping integrals $t$ has $E(\mathbf{k})=E_0+2t{\sum }_{i=1}^3\cos (k_{i}a)$ and $W=12|t|$.

The interplanar spacing $d$ is the perpendicular distance between parallel planes in a specified family. Measuring along an oblique line gives a larger distance, so the direction of measurement matters.

There is a short reciprocal lattice derivation. With the angular convention $a_i\cdot b_j=2\pi {\delta }_{ij}$, put $G=h{b}_1+k{b}_2+l{b}_3$. Then $G\cdot r=2\pi m$ is the same plane equation as $hx+ky+lz=m$. Thus $G$ is a normal vector. Moving between consecutive sheets along this normal changes the dot product by $2\pi$, givingFor perpendicular cubic axes of length $a$, this becomesFor example, choose a primitive cubic lattice with $a=0.600\mathrm{nm}$. Its $(210)$ lattice planes have $d_{210}=0.600/\sqrt{5}\simeq 0.268328\mathrm{nm}$. This is a chosen geometry, not an experimental value. The distance from a point to a plane gives the same result: the planes $2X+Y=0$ and $2X+Y=a$ in ordinary Cartesian length coordinates are separated by $a/\sqrt{5}$.

Specify whether $d_{hkl}$ denotes spacing between successive occupied lattice planes or the formal spacing assigned to a reflection label. In a primitive cubic lattice, $(200)$ has formal spacing $a/2$ but every other sheet contains no lattice points; its reflection is the second diffraction order of $(100)$. In a face-centered cubic lattice with conventional cubic axes, the $(200)$ sheets at half-integer $X/a$ do contain lattice points. Here $a/2$ is the actual spacing of the complete family perpendicular to that cube edge. The Primitive cubic Bragg peak positions example explains how to avoid counting a reflection twice.

Choose an original simulated model: a primitive cubic lattice with one identical point scatterer per primitive unit cell, edge $a=0.400\mathrm{nm}$, and incident X-ray wavelength $\lambda =0.200\mathrm{nm}$. These are stipulated lengths, not measurements of sodium chloride. Suppose orientation information has already assigned the two observed reflections the Miller indices $(100)$ and $(110)$. The detector deflection is $2\theta$, where $\theta$ is the Bragg angle.

Use the existing interplanar spacing formula and Bragg's law, with higher diffraction orders absorbed into the reflection indices:The simulated $(100)$ reflection has $\sin \theta =1/4$, hence $2\theta =28.9550^{\circ }$. The $(110)$ reflection has $\sin \theta =\sqrt{2}/4$, hence $2\theta =41.4096^{\circ }$. Their squared sines are $0.0625$ and $0.125$, in ratio $1:2$. Substituting the unrounded angles independently givesAgreement tests a proposed indexing and scale together; disagreement would prompt checks of the indexing, wavelength, angular zero, or the assumed unit cell geometry.

Two peak positions alone do not establish the indexing. Assigning the same two angles to $(110)$ and $(200)$ would give $N=2,4$ and $a=0.400\sqrt{2}\simeq 0.565685\mathrm{nm}$ instead. More reflections, orientation information and systematic absences help discriminate such assignments. An unobserved reflection can be inaccessible, weak or extinguished by the atomic basis; absence is not automatically evidence for a larger interplanar spacing.

Absolute scale also requires measurement calibration of the wavelength. To first order, differentiation of the inferred length giveswith angular changes in radians. A stipulated detector-zero error $\delta (2\theta )=+0.100^{\circ }$, with correct wavelength, biases the first inferred edge downward by approximately $0.338\%$ and the second by $0.231\%$. A common angular offset therefore need not cancel when two inferred lengths are compared. These are model sensitivity estimates, not historical instrument uncertainties.

A lattice plane is a plane containing a two-dimensional periodic array of lattice points. Parallel lattice planes form equally spaced geometrical sheets; an atomic basis can place chemical atoms between them.

With primitive lattice translation vectors $a_i$, write $r=x{a}_1+y{a}_2+z{a}_3$. For integers $h,k,l$ with greatest common divisor one, the plane equationsgive a complete family labeled by Miller indices $(hkl)$. In a primitive cubic lattice, $(110)$ means $x+y=m$, parallel to $a_3$.^[ref]

In the Laue method, a fixed single crystal structure receives a collimated beam containing many wavelengths. Each reciprocal lattice vector can select its own wavelength; waves with different wavelengths need not interfere with one another. For $\mathbf{G}\ne 0$, the elastic scattering constraint gives the particularly useful testA candidate reflection requires ${\lambda }_{\mathbf{G}}>0$ within the source spectrum and an outgoing ray that reaches the detector. Varying wavelength changes the radius and center of the Ewald sphere while leaving the reciprocal lattice fixed. A monochromatic wave instead fixes the radius: rotating the crystal structure or varying the incident wavelength in a scan finds reflections sequentially. A white-beam photograph integrates many spectral selections in one exposure; it is not an angular scan at one known wavelength.

The Laue method is useful for orienting a single crystal structure: tilting it changes which reciprocal lattice vectors meet the spectral and detector conditions. Absolute scale has an ambiguity if the wavelengths are unknown. Scaling all direct lattice translations by $c>0$ and all selected wavelengths by $c$ leaves every outgoing direction unchanged. Independently known spectral information or length calibration is needed to remove this ambiguity.

Ordinary solid sodium chloride has the rock salt structure. Choose a conventional cube of side $a$ with chlorine anions at the corners and face centers, and sodium cations at the edge centers and cube center. Corner sites are shared by eight cubes, face sites by two and edge sites by four. The unit cell counts are thereforeThere are four formula units per conventional unit cell, rather than four separate NaCl molecules.^[ref]

The Bravais lattice is a face-centered cubic lattice. One choice of primitive lattice translation vectors and atomic basis isThe determinant gives a primitive unit cell volume of $a^3/4$; that cell represents one cation and one anion. Four such volumes make the conventional cube. The union of all ion positions resembles a primitive cubic lattice of step $a/2$ only if the chemical labels are discarded: translation by $a(1,0,0)/2$ exchanges unlike ions and fails to preserve the chemical structure.

Each ion has six nearest unlike neighbors at distance $a/2$, along the positive and negative directions of the three cube axes. Thus the coordination number is six. Within either same-species face-centered cubic lattice, the nearest same-ion neighbors instead number twelve: their displacements are permutations of $(\pm a/2,\pm a/2,0)$, all of length $a/\sqrt{2}$. Since $a/\sqrt{2}>a/2$, those twelve are a more distant shell in the complete sodium chloride structure. The number twelve describes the same-ion FCC lattice neighborhood; six describes the nearest unlike-ion neighborhood.

For a spherical Fermi surface inside a primitive cubic lattice zone, the shortest reciprocal vector has magnitude $G=2\pi /a$. The nearest translated Fermi sphere is separated by the electron-phonon Umklapp thresholdA phonon of this wavevector can connect nearly opposite-velocity Fermi states using $\mathbf{k}+\mathbf{q}={\mathbf{k}}^{\prime }+\mathbf{G}$, transferring reciprocal-lattice momentum to the crystal and efficiently relaxing electrical current. With the supplied sound speed, its thermal scale isFor $T\ll T_U$ such phonons have exponentially small occupation. As they become populated, Umklapp scattering can make the electrical conductivity fall rapidly. Lower-energy normal scattering and residual disorder still give resistance below this scale; this is a crossover estimate, not a sharp transition. The numerical estimate extrapolates the linear acoustic dispersion to $q_U$ and assumes the spherical Fermi surface implicit in giving one $k_F$.

Using $2k_F$ alone as the phonon wavevector misses this mechanism: direct backscattering would need $1.4\pi /a$, beyond the first-zone edge along this direction, while the reciprocal-lattice transfer permits it with $0.6\pi /a$.

For state counting in a Bloch band, take a primitive cubic lattice of spacing $a$ and impose periodic boundary conditions on lengths $L_i=N_{i}a$. The allowed wavevectors are $k_i=2\pi l_i/L_i$. The first Brillouin zone has side $2\pi /a$, so it contains $N_i$ distinct values in each direction and hence $N_x{N}_y{N}_z=N$ wavevectors. Each orbital-derived energy band has one state per wavevector per spin orientation. ThereforeThe phrase “per atom” attached to $2N$ in the PDF is a counting error: $2N$ is the total, not the per-atom count.

By the Pauli exclusion principle, a monovalent crystal supplies $N$ valence Electrons and half-fills a spin-degenerate band; its Fermi level lies within allowed states, giving a metal in the independent-electron picture. Two electrons per atom can fill one band. If a band gap separates it from all empty bands, the crystal is a band insulator. An even valence count only permits insulation: overlapping bands can still produce a metal. Conversely, electron interactions can produce a Mott insulator, and symmetry breaking can permit insulation outside this simple spin-degenerate electronic band structure criterion. The relevant count in a crystal with a basis is per primitive unit cell.

A site in a primitive cubic lattice has six nearest neighbours, at $\pm a\hat{\mathbf{x}}$, $\pm a\hat{\mathbf{y}}$ and $\pm a\hat{\mathbf{z}}$. Repeating the Bloch state calculation with equal hopping integrals $t$ along all six bonds givesThe cosine sum ranges from $-3$ to $3$ within the first Brillouin zone. ThereforeFor $t<0$ the minimum is at the zone centre; for $t>0$ the maximum is there.

Consider a chosen primitive cubic lattice $a{\mathbb{Z}}^3$ with $a=0.400\mathrm{nm}$ and one identical point scatterer per primitive unit cell. Retain the incident wavelength $\lambda =0.154\mathrm{nm}$. This is an ideal point model. Its reciprocal lattice vectors are $q=(2\pi /a)(h,k,l)$ for integer triples other than $(0,0,0)$. Write $S=h^2+k^2+l^2$. The associated interplanar spacing and first-order Bragg angle satisfyThe reflection labels $(hkl)$ use the Miller index convention, keeping unreduced triples to encode diffraction order. With that convention, set $n=1$ in Bragg's law for each label. The $(200)$ reflection is equivalently the second diffraction order of $(100)$: $d_{200}=a/2$ does not imply extra scatterers halfway between the occupied $(100)$ planes. Counting both descriptions as distinct reflections would count the same direction twice.

For $(100)$, $(110)$, $(111)$ and $(200)$, respectively, $S=1,2,3,4$ and the calculated interplanar spacings are $0.400000$, $0.282843$, $0.230940$ and $0.200000\mathrm{nm}$. Their calculated deflections $2\theta$ are $22.197435^{\circ }$, $31.594744^{\circ }$, $38.952948^{\circ }$ and $45.287481^{\circ }$. Equal steps in $S$ give equal steps in ${\sin }^2\theta$, not equal steps in $2\theta$. This regularity makes an angular pattern a test of a proposed crystal structure.

A real crystal structure has a Bravais lattice together with an atomic basis of a crystal. The crystal lattice structure factor selects the periodic directions; the crystal basis structure factor combines the scattering amplitudes of atoms within a unit cell and can cancel a reflection. Consequently Bragg's law alone predicts neither allowed reflections nor their wave intensities. ^[ref] Systematic absences further restrict observed families; for example, face-centered cubic lattice centering permits only indices all even or all odd in conventional cubic indexing. The four strips above apply to the stated primitive cubic lattice, not to sodium chloride or a face-centered cubic lattice. ^[ref]