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The Bragg angle is the glancing angle from the incident beam to the selected lattice planes, with $0<\theta \le 90^{\circ }$. If $\alpha$ is instead the angle from the beam to the plane normal, $\alpha =90^{\circ }-\theta$ and Bragg's law becomes $n\lambda =2d\cos \alpha$. Defining the angle explicitly prevents a sine-versus-cosine error as well as a factor-of-two error in the detector deflection.

Choose $\lambda =0.154\mathrm{nm}$ and $d=0.200\mathrm{nm}$ as an illustrative model, not historical measurements. For the first diffraction order,The second diffraction order has $\sin \theta =0.770$ and $2\theta \simeq 100.707778^{\circ }$; the third would require $\sin \theta =1.155$ and is impossible. These are different diffraction orders of the same interplanar spacing at the same wavelength.

Conversely, a first-order reflection at a measured deflection $2\theta$ gives the interplanar spacingUsing $2\theta \simeq 45.287481^{\circ }$ recovers $d\simeq 0.200\mathrm{nm}$. This is the useful reversal: a known wavelength turns an angular measurement into an atomic length. An unassigned diffraction order or an unknown wavelength prevents that single measurement from determining $d$ uniquely.

Compare two rays scattered into parallel outgoing directions from neighboring lattice planes. In the diagram, $A$ lies on the upper plane and $B$ on the lower plane, directly beneath $A$ with $AB=d$. Draw through $A$ the incoming and outgoing equal-wave phase fronts, perpendicular to their respective rays. They meet the lower ray at $C$ and $D$. Each right triangle gives $CB=BD=d\sin \theta$, so the lower ray travels the extra distanceThe resulting phase difference is $\Delta \phi =2\pi \Delta \ell /\lambda$. Constructive interference requires $\Delta \phi =2\pi n$, which proves Bragg's law in this ray construction. Comparing equal-wave phase fronts, rather than arbitrary ray endpoints, is what makes the distance comparison meaningful.

The same geometry connects to the reciprocal lattice. With wavenumber $k=2\pi /\lambda$, the elastic Bragg scattering condition gives a transfer magnitude $|q|=2k\sin \theta$. A reciprocal vector normal to these lattice planes has $|q|=2\pi n/d$. Equating them again yields Bragg's law. The ray and reciprocal lattice descriptions therefore express the same wave phase matching.

Bragg's law selects directions of constructive interference in coherent elastic scattering from equally spaced lattice planes:Here $d>0$ is the perpendicular interplanar spacing, $\lambda >0$ the incident wavelength, $n$ the positive integer diffraction order, and $\theta$ the Bragg angle, measured between the incoming beam and the lattice planes. The scattering deflection, between incoming and outgoing propagation directions, is $2\theta$. Thus a plotted detector angle $2\theta$ must be halved before insertion into Bragg's law. The $\theta$ in the parent elastic Bragg scattering condition denotes the full deflection; the present $\theta$ denotes half of it. ^[ref]

The lattice planes organize wave interference throughout a crystal structure; they are not literal metal mirrors or separate reflecting surfaces. The picture assumes elastic scattering, hence an unchanged wavelength, and sufficiently stable relative wave phases across the contributing lattice planes. X-ray wavelengths comparable with atomic interplanar spacings make these directions experimentally accessible. The sine cannot exceed one: an order exists geometrically only if $n\lambda /(2d)\le 1$. For $\lambda >2d$, even the first diffraction order is inaccessible. A geometrically possible direction need not have nonzero wave intensity; the crystal lattice structure factor and crystal basis structure factor also matter. ^[ref]

A crystal X-ray spectrometer selects a narrow wavelength interval by diffraction at known interplanar spacing $d$ and Bragg angle $\theta$. For fixed diffraction order $n$, a known $d$ lets an angular measurement determine $\lambda =2d\sin \theta /n$. This reverses the crystallography problem of finding $d$ from a known wavelength.

Siegbahn's laboratory developed X-ray tubes and evacuated spectrographs for longer wavelengths. Rotating the crystal selected reflections; photographic plates recorded positions. Opposite-side exposures measured angles on both sides of the direct beam. Later tube instruments mounted a beam-defining slit opposite the plate holder. A separate double instrument compared two reflected beams with two ionization chambers. These were different instruments, not one universal assembly.^[ref]

The Nobel biography describes constant-temperature spectrographs and a precision angular scale read by diametrically opposed microscopes.^[ref] Temperature control matters because thermal expansion changes $d$; an accurately read angle cannot compensate for an unknown interplanar spacing.

The schematic separates preparation, selection and readout. An X-ray tube target supplies Characteristic X-rays with a possible bremsstrahlung background. A collimator limits incident directions; evacuation reduces air X-ray attenuation. Rotating the crystal brings different wavelengths into reflection. A photographic plate stores line positions over an exposure, whereas an ionization chamber can measure collected electric charge at successive angles. Neither readout alone establishes absolute wavelength: an angular zero, angular-scale calibration, diffraction order, $d$ and any required refraction correction enter the inference. Narrower slits reduce accepted angular spread but also reduce signal; a weak line can become harder to locate despite a narrower instrumental response.

A control reported in Siegbahn's lecture photographed a line system with copper-sulphate and calcium-sulphate crystals. Irregular relative shifts around copper's characteristic frequency supported anomalous dispersion; the arrangement excluded differing absorption across that frequency as the explanation.^[ref] A finer angular scale alone cannot remove this physical refraction correction.

Let an Electron start with negligible kinetic energy and move through an increase $U>0$ in electric potential. Its electric charge is $q=-e$, where $e>0$ is the elementary charge, so the decrease in potential energy is $eU$. In a field-free region after acceleration,This is an Electron formula, not a universal conversion from voltage to wavelength. It assumes negligible initial kinetic energy, negligible losses, and $eU\ll m_e{c}^2$, where $c$ is the speed of light. It describes the de Broglie wavelength outside the crystal; the potential energy inside a material can change the local kinetic momentum.

For $U=54\mathrm{V}$, take $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $e=1.602176634\times 10^{-19}\mathrm{C}$, and $m_e\simeq 9.1093837\times 10^{-31}\mathrm{kg}$. The rounded Electron mass is consistent with the 2022 recommended constants.^[ref]The electronvolt is an energy unit: $54\mathrm{eV}$ must be converted to joules before combining it with $m_e$ in SI units. Dimensional analysis checks both steps:The resulting velocity is about $4.3584\times 10^6\mathrm{m}{\mathrm{s}}^{-1}$, or $0.01454c$, so the nonrelativistic assumption is consistent. For a larger accelerating voltage, the relativistic energy-momentum relation gives $pc=\sqrt{K(K+2m_e{c}^2)}$, with $K=eU$, while $\lambda =h/p$ still holds. At $54\mathrm{V}$ the relativistic wavelength is shorter by about $0.00264\%$.

Explicitly, the exact special relativity formula and its nonrelativistic quantum mechanics limit areHere $K=eU$ is the gained kinetic energy; the total energy is $E=m_e{c}^2+K$. At $80\mathrm{kV}$, the nonrelativistic quantum mechanics expression gives $4.336\mathrm{pm}$, about $3.84\%$ larger than the exact $4.176\mathrm{pm}$. This difference matters when using electron diffraction ring positions to infer lattice plane spacings.

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.

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]

For incident and outgoing unit propagation vectors ${\mathbf{s}}_{\mathrm{in}}$ and ${\mathbf{s}}_{\mathrm{out}}$, wavelength $\lambda >0$, and primitive lattice translation vectors $a_1,a_2,a_3$, the Laue equations areAll three must hold for the same wavelength and outgoing direction. The integers label a reciprocal lattice vector; they need not be coprime. For wavevectors ${\mathbf{k}}_{\mathrm{in},\mathrm{out}}=(2\pi /\lambda ){\mathbf{s}}_{\mathrm{in},\mathrm{out}}$, equivalentlyThis uses angular wavenumbers; the crystallographic convention without $2\pi$ instead uses $\mathbf{s}/\lambda$.^[ref]

For a short proof, translate a scatterer through $a_i$. Its incident wave phase changes by ${\mathbf{k}}_{\mathrm{in}}\cdot a_i$ and propagation to the same distant observation direction contributes $-{\mathbf{k}}_{\mathrm{out}}\cdot a_i$. The relative wave phase factor is therefore $\exp (-i\mathbf{G}\cdot a_i)$. It equals one for all lattice translations exactly when $\mathbf{G}$ belongs to the reciprocal lattice. A finite crystal lattice structure factor broadens the ideal maxima; the crystal basis structure factor can cancel them. Geometric phase matching is necessary for a sharp reflection, but does not guarantee a visible spot.

For elastic scattering, the additional constraint isIt is an Ewald sphere centered at $-{\mathbf{k}}_{\mathrm{in}}$ in the space of transfer vectors $\mathbf{G}$. The parent elastic Bragg scattering condition uses $q={\mathbf{k}}_{\mathrm{in}}-{\mathbf{k}}_{\mathrm{out}}=-\mathbf{G}$, so its displayed condition has the opposite sign. Both give Bragg's law for the same lattice planes.

Three Miller indices $(hkl)$ specify a lattice plane orientation in chosen unit cell axes. Choose a parallel plane away from the origin, express its intercepts in axis units, take reciprocals, and clear fractions. An infinite intercept gives index zero; a negative intercept gives a negative index, written with an overbar.^[ref]

For example, intercepts $(2a,3b,\infty )$ have reciprocal ratios $(1/2,1/3,0)$, so the orientation is $(320)$ after multiplying by six. Relative to the same axes, intercepts $(a,-b,\infty )$ give $(1\bar{1}0)$. Choosing a different parallel member changes the common intercept scale, not the orientation.

Primitive axes require relatively prime Miller indices for complete families. Centered conventional unit cells need not, as the FCC lattice $(200)$ example shows. Reflection labels can also retain higher diffraction orders. Use the reciprocal lattice for the normal vector with oblique axes; the direct vector $h{a}_1+k{a}_2+l{a}_3$ need not be perpendicular.

In neutron diffraction, coherent elastic scattering adds scattering amplitudes from regularly positioned scatterers. Since $E_f=E_i$, the neutron scattering kinematics gives $k_f=k_i=2\pi /\lambda$ andFor lattice planes of interplanar spacing $d_{\mathrm{p}}$, constructive addition requires $Q=2\pi n/d_{\mathrm{p}}$ along the plane normal. Equating these expressions proves Bragg's law:For example, stipulate $d_{\mathrm{p}}=2$ ångströms and $\lambda =2$ ångströms. The first diffraction order has $\theta =30^{\circ }$ and detector deflection $60^{\circ }$. A peak position constrains a period; its intensity also depends on the crystal basis structure factor.

Clifford Glenwood Shull describes powdered sodium chloride, a crystal monochromator, and a shielded counter recording angular counts. Inserting cadmium measured background; substantial diffuse scattering complicated peak interpretation.^[ref]

Powdering supplies many crystal orientations. Comparing the angular peaks with Bragg's law constrains crystal structure; a background estimate helps distinguish specimen peaks from unwanted counts. A two-axis diffraction instrument has no separate outgoing energy analysis, so an angular pattern alone does not separate every inelastic contribution. The richer magnetic example, including a warm/cold comparison and its limits, is the manganese monoxide neutron diffraction experiment. Its extra peaks reveal magnetic order beyond the chemical periodicity; the associated magnetic neutron diffraction model is developed there.

Alongside the 1927 Davisson–Germer experiment, George Paget Thomson developed an independent transmission test. Alexander Reid first used celluloid; Thomson used thin metal films. A narrow Electron beam crossed a film about $10^{-6}\mathrm{cm}$ thick, reaching a photographic plate perpendicular to the beam. Rings appeared. Accelerating voltages of $20$–$60\mathrm{kV}$ set the de Broglie wavelengths; metal structures known from X-ray diffraction allowed quantitative comparison. A downstream magnetic field deflected ring-forming Electrons like those passing through holes, identifying the carriers as Electrons rather than X-rays. Celluloid's uncertain structure limited its quantitative interpretation.^[ref]

The Davisson–Germer experiment uses crystal reflection and angular electric current scans; this transmission camera records a spatial pattern behind the specimen. Both probe interference in electron diffraction, with different specimen preparations and readouts. A developed photographic plate records accumulated detections, not a direct image of the wavefunction.

Why rings? In an ideal film containing many small crystals with uniformly distributed orientations, a family of lattice planes with spacing $d$ can satisfy Bragg's law at every azimuth about the incident beam. First-order elastic scattering then forms a cone of opening half-angle $2\theta$, where $2d\sin \theta =\lambda$. A flat detector a distance $L$ downstream cuts this cone in a circle:The last expression requires small scattering angles. It explains a useful control: hold the specimen and camera distance fixed, increase accelerating voltage, and the ring radii should shrink with the electron accelerating-voltage wavelength. It predicts positions, not intensities: preferred crystal orientations, film thickness and repeated quantum scattering can alter a real pattern.

For a chosen example, take $L=0.200\mathrm{m}$ and $d=0.200\mathrm{nm}$. The exact relativistic electron accelerating-voltage wavelength gives $\lambda =8.589\mathrm{pm}$ at $20\mathrm{kV}$ and $4.176\mathrm{pm}$ at $80\mathrm{kV}$. The corresponding exact ring radii are $8.594\mathrm{mm}$ and $4.176\mathrm{mm}$. These are calculated examples with chosen spacings and camera distance, not Thomson's measurements. The higher voltage lies outside the range quoted from his lecture, deliberately illustrating a prediction rather than reconstructing his data.