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The de Broglie wavelength is the distance over which a matter wave accumulates one full spatial quantum phase cycle in its propagation direction:Here $h$ is the Planck constant, $\mathbf{p}$ is the momentum, and $\mathbf{k}$ is the angular wavevector. The scalar $p$ in $\lambda =h/p$ means the positive momentum magnitude, not a signed component. Reversing a momentum reverses the wavevector without making the wavelength negative. At zero momentum there is no finite spatial phase period: the limiting wavelength diverges.

This relation makes microscopic diffraction accessible with particles: increasing the momentum shortens the wavelength until it is comparable with the spacings in a crystal scattering problem. The electron accelerating-voltage wavelength gives a practical way to tune this scale, while the Davisson–Germer experiment tests it by comparing electron diffraction directions with independently known crystal geometry. The same relation applies to neutral particles in a neutron interferometer; electric charge is not a requirement for a matter wave.

The wavelength belongs to the wavefunction, not to a sinusoidal path of a particle. A free-particle momentum convention makes the distinction especially clear: a plane wave has a rotating complex quantum phase but a constant probability density. A localized wave packet contains a range of momenta and therefore generally has a range of wavelengths, rather than one exact spatial period.

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.

Electron microscope resolution concerns distinguishable specimen detail under a specified contrast criterion. It depends on diffraction, lens defects, electron energy spread, specimen motion and detection. A smaller de Broglie wavelength permits finer detail in principle, but neither wavelength nor magnification is the instrument's measured resolution.

For an electron accelerated from negligible initial kinetic energy through a positive electric potential difference $U$, combine $K=eU$, the relativistic energy-momentum relation and the de Broglie wavelength:Here $h$ is the Planck constant, $m_e$ the electron mass, $e$ the elementary charge, and $c$ the speed of light. This includes the relativistic correction omitted by the low-voltage electron accelerating-voltage wavelength approximation. With $m_e\simeq 9.1093837\times 10^{-31}\mathrm{kg}$ and SI defining values of $h,e,c$, calculated wavelengths are approximately $6.979$, $3.701$ and $1.969\mathrm{pm}$ at $30$, $100$ and $300\mathrm{kV}$ respectively. The rounded mass agrees with the recommended value.^[ref]

For a circular aperture admitting a small semi-angle $\alpha$, the Rayleigh criterion suggests a diffraction-limited resolution scale $d_D=0.61\lambda /\alpha$. At $100\mathrm{kV}$ and $\alpha =10\mathrm{mrad}$, this gives $0.226\mathrm{nm}$, already much larger than the $0.00370\mathrm{nm}$ wavelength. This is an ideal aperture estimate. Transmission electron microscopy with coherent illumination and a particular specimen requires its own image-contrast analysis; this scale alone is not a universal two-feature resolution test.

An original illustrative blur model adds competing effects:Define $C_s{\alpha }^3$ as a characteristic transverse blur from the electron-lens analogue of spherical aberration, and $C_c\alpha \Delta K/K$ as one from energy-dependent focusing, the analogue of chromatic aberration. The coefficients $C_s,C_c$ have units of length, $\Delta K$ is a chosen characteristic kinetic energy spread, and $b_0$ is a fixed residual blur. Adding their squares is a declared heuristic, not a measured point spread function or a rigorous resolution theorem. The model omits contrast transfer, specimen scattering and signal statistics, so its minimum is only an estimate within its assumptions.

Choose $C_s=C_c=1\mathrm{mm}$, $\Delta K=0.7\mathrm{eV}$ and $b_0=0.10\mathrm{nm}$ at $K=100\mathrm{keV}$. At $5\mathrm{mrad}$, the four contributions are about $0.452$, $0.125$, $0.035$ and $0.100\mathrm{nm}$, giving $b\simeq 0.480\mathrm{nm}$. At $10\mathrm{mrad}$, the diffraction contribution falls to $0.226\mathrm{nm}$ but the assumed spherical aberration contribution grows to $1.00\mathrm{nm}$. Thus opening the aperture indefinitely worsens this model's image. All these values are calculations with chosen parameters, not observations of any microscope.

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.