AcademyMatter Waves
Academy
Electron Waves
Level 1 - Physics topic page in Matter Waves.
Principle
Particles with momentum have de Broglie wavelengths and can show interference and diffraction.
Notation
\(\lambda\)
de Broglie wavelength
\(\mathrm{m}\)
\(h\)
Planck constant
\(\mathrm{J\,s}\)
\(p\)
momentum magnitude
\(\mathrm{kg\,m\,s^{-1}}\)
\(m_e\)
electron mass
\(\mathrm{kg}\)
\(e\)
elementary charge
\(\mathrm{C}\)
\(V\)
accelerating potential difference
\(\mathrm{V}\)
Method
Derivation 1: Attach a wavelength to momentum
The de Broglie relation assigns a wave scale to any particle with momentum.
de Broglie relation
\[\lambda=\frac{h}{p}\]
Nonrelativistic momentum
\[p=mv\]
Derivation 2: Use an accelerating voltage
An electron accelerated from rest through \(V\) gains kinetic energy \(eV\).
Energy gain
\[K=eV\]
Momentum from energy
\[p=\sqrt{2m_eK}=\sqrt{2m_eeV}\]
Electron wavelength
\[\lambda=\frac{h}{\sqrt{2m_eeV}}\]
Derivation 3: Read diffraction as wave evidence
If \(\\lambda\) is comparable with atomic spacing, electrons diffract from crystals like waves from a grating.
Rules
de Broglie
\[\lambda=\frac{h}{p}\]
Electron energy
\[K=eV\]
Electron wavelength
\[\lambda=\frac{h}{\sqrt{2m_eeV}}\]
Bragg condition
\[2d\sin\theta=n\lambda\]
Examples
Question
An electron has momentum
\[2.00\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}\]
Find its de Broglie wavelength.Answer
\[\lambda=\frac{h}{p}=\frac{6.63\times10^{-34}}{2.00\times10^{-24}}=3.32\times10^{-10}\,\mathrm m\]
Checks
- Larger momentum means shorter wavelength.
- Use joules when substituting into \(\lambda=h/p\).
- Electron diffraction is strongest when wavelength and spacing are comparable.
- The voltage formula is nonrelativistic; high voltages need relativistic momentum.