Calculate the de Broglie wavelength of a 1.68 keV electron
Use this premium calculator to find the de Broglie wavelength for an electron with kinetic energy of 1.68 keV. You can switch units, compare nonrelativistic and relativistic models, and visualize how wavelength changes with electron energy.
Calculator Inputs
Default example: 1.68 keV electron. This tool applies the de Broglie relation λ = h/p, with an optional relativistic momentum correction.
Results
Wavelength vs electron energy
Expert guide: how to calculate the de Broglie wavelength of a 1.68 keV electron
The de Broglie wavelength is one of the central ideas in quantum mechanics. It expresses the wave like behavior of matter by assigning a wavelength to any moving particle. For an electron, this concept is especially important because electrons are light, fast, and commonly used in microscopy, diffraction, spectroscopy, and solid state physics. If you want to calculate the de Broglie wavelength of a 1.68 keV electron, the process is straightforward once you identify the correct formula and keep your units consistent.
In this page, the goal is very specific: determine the wavelength associated with an electron whose kinetic energy is 1.68 keV. Since 1 keV equals 1000 electron volts, the input energy is 1680 eV. At this energy, the electron is still much lower than its rest energy of about 511 keV, which means the nonrelativistic formula already gives a very good approximation. Even so, a relativistic calculation is slightly more precise, and this calculator can show both.
What the de Broglie wavelength means
Louis de Broglie proposed that particles can exhibit wave behavior, with wavelength given by:
λ = h / p
Here, λ is the wavelength, h is Planck’s constant, and p is the particle momentum. For photons, wave behavior was already well established. De Broglie’s major insight was that material particles such as electrons, neutrons, and atoms should also have wavelengths. This prediction was later confirmed experimentally, most famously by electron diffraction. Today, the de Broglie wavelength is essential in understanding why electrons produce diffraction patterns in crystals and why electron microscopes can resolve fine structures much smaller than visible light can.
Step by step calculation for a 1.68 keV electron
- Convert the energy to electron volts if needed. Here, 1.68 keV = 1680 eV.
- Choose a model: nonrelativistic or relativistic.
- Find the electron momentum from kinetic energy.
- Apply λ = h/p.
- Convert the result into practical units such as picometers or angstroms.
In the nonrelativistic approximation, the kinetic energy is:
K = p2 / 2m
Solving for momentum gives:
p = √(2mK)
Substituting into the de Broglie relation:
λ = h / √(2mK)
If you use SI units, you need:
- Planck’s constant h = 6.62607015 × 10-34 J·s
- Electron mass m = 9.1093837015 × 10-31 kg
- 1 eV = 1.602176634 × 10-19 J
Now convert 1680 eV to joules:
K = 1680 × 1.602176634 × 10-19 J = 2.69165674512 × 10-16 J
Then compute momentum:
p = √(2mK)
which gives a momentum of about 2.214 × 10-23 kg·m/s, and therefore:
λ ≈ 2.993 × 10-11 m
That is:
- 0.02993 nm
- 29.93 pm
- 0.2993 Å
Why the relativistic correction matters
Even though 1.68 keV is far below 511 keV, the electron is still moving fast enough that a relativistic formula is slightly better. For a relativistic particle, the momentum can be obtained from:
(pc)2 = K(K + 2mc2)
So the wavelength becomes:
λ = hc / √(K(K + 2mc2))
For electrons, it is often convenient to use energies in electron volts directly, since both kinetic energy and rest energy can be written in eV. Using m c2 = 510998.95 eV and K = 1680 eV, the wavelength comes out to roughly:
λ ≈ 2.9905 × 10-11 m = 29.905 pm = 0.29905 Å
The difference between nonrelativistic and relativistic answers is small, around a few hundredths of a picometer. In introductory work, that difference is usually negligible. In high precision beam physics, diffraction calibration, or instrument design, it becomes meaningful.
| Method | Formula used | Result for 1.68 keV electron | Practical interpretation |
|---|---|---|---|
| Nonrelativistic | λ = h / √(2mK) | 29.929 pm | Excellent approximation at low electron energies |
| Relativistic | λ = hc / √(K(K + 2mc²)) | 29.905 pm | Preferred for more accurate calculations |
| Difference | Approximation error | 0.024 pm | Small, but visible in precision comparison |
Useful shortcut formula for electrons
For electrons accelerated through a potential difference V in volts, a widely used shortcut is:
λ(Å) ≈ 12.27 / √V
That is the nonrelativistic expression. Since a 1.68 keV electron corresponds to 1680 V of accelerating potential, you get:
λ ≈ 12.27 / √1680 ≈ 0.299 Å
A more accurate relativistic version is:
λ(Å) ≈ 12.27 / √(V(1 + 0.978 × 10-6V))
With V = 1680, that yields nearly the same result but with the expected slight downward correction. This is why many lab manuals and microscopy references quote electron wavelengths directly from accelerating voltage rather than from a separate kinetic energy conversion.
How small is 29.9 pm?
A picometer is one trillionth of a meter. A wavelength near 30 pm is much smaller than visible light, whose wavelengths are roughly 380 to 750 nm. This enormous difference explains why electron beams can probe much finer details than optical microscopes. It also explains why electrons readily diffract from atomic spacings in solids, which are often on the order of 100 to 300 pm.
| Wave or length scale | Typical value | Comparison with 1.68 keV electron wavelength |
|---|---|---|
| 1.68 keV electron de Broglie wavelength | 29.9 pm | Reference value |
| Typical carbon carbon bond length | 154 pm | About 5.1 times larger |
| Silicon lattice constant | 543 pm | About 18.2 times larger |
| Visible green light | 532000 pm | About 17790 times larger |
Applications of a 1.68 keV electron wavelength
A wavelength around 30 pm places the electron solidly in the range where atomic scale diffraction is possible. This matters in several domains:
- Electron diffraction: electrons scatter from crystal planes, producing patterns that reveal lattice spacing and symmetry.
- Surface science: low energy electron techniques are often sensitive to top surface layers and reconstructions.
- Quantum mechanics education: this energy range is excellent for demonstrating matter wave concepts in a manageable regime.
- Instrumentation: estimating beam wavelength is useful when discussing resolution limits, coherence, and scattering behavior.
It is important to remember that wavelength alone does not determine image quality or measurement resolution. Real systems also depend on aberrations, coherence, detector quality, interaction volume, sample thickness, and environmental stability. Still, wavelength sets a hard physical scale and remains one of the first numbers scientists calculate when characterizing an electron beam.
Common mistakes when calculating electron wavelength
- Forgetting unit conversion: keV must be converted correctly to eV or joules.
- Mixing formulas: use the nonrelativistic formula only when kinetic energy is well below the rest energy.
- Confusing energy and voltage: for a singly charged electron accelerated through V volts, the kinetic energy is eV, numerically equal to V electron volts.
- Using the wrong particle mass: the electron mass must be used, not proton or neutron mass.
- Reporting awkward units: in practice, pm or Å are usually much easier to interpret than meters.
Interpreting the answer in context
When you calculate the de Broglie wavelength of a 1.68 keV electron and obtain about 29.9 pm, you are measuring the scale of the electron’s quantum phase variation in free space. This does not mean the electron is literally a tiny sinusoidal string. Instead, it means the electron’s wavefunction has a spatial periodicity tied to its momentum. When that periodicity is comparable to structural distances in a material, interference and diffraction become observable. That is why the de Broglie relation is so powerful: it connects particle motion with wave phenomena in a precise quantitative way.
In many textbooks, this calculation serves as a bridge between classical and quantum ideas. You start from an electron with a known kinetic energy, use mechanical relations to find momentum, and then use de Broglie’s quantum relation to assign a wavelength. The result is immediately useful experimentally, because it predicts diffraction conditions and approximate resolving power.
Recommended references and authoritative sources
- NIST Fundamental Physical Constants
- U.S. Department of Energy overview of quantum mechanics
- Georgia State University HyperPhysics on de Broglie wavelength
Final answer
If you need the direct result only, here it is:
- Nonrelativistic de Broglie wavelength: approximately 29.929 pm
- Relativistic de Broglie wavelength: approximately 29.905 pm
So, the de Broglie wavelength of a 1.68 keV electron is best reported as about 29.9 pm, which is also 0.299 Å or 2.99 × 10-11 m.