The formulas
- de Broglie wavelength: λ = h/p = h/(mv)
- From kinetic energy: λ = h/√(2mK)
- Charged particle through a voltage V: K = qV, so λ = h/√(2mqV)
- Electron shortcut: λ = 12.27/√V Å (or 1.227/√V nm)
- Proton: λ = 0.286/√V Å. Alpha particle: λ = 0.101/√V Å
Electron wavelength at common voltages
| Voltage (V) | λ (Å) |
|---|---|
| 1 | 12.3 |
| 10 | 3.88 |
| 50 | 1.74 |
| 100 | 1.23 |
| 150 | 1 |
| 500 | 0.549 |
| 1000 | 0.388 |
| 10000 | 0.123 |
Why don't we see a cricket ball diffract? A 0.16 kg ball at 40 m/s has λ = 6.6 × 10⁻³⁴ ÷ 6.4 ≈ 10⁻³⁴ m, far smaller than anything it could pass through. Choose "Other" and try it.
Photon energy from wavelength is on the Rydberg formula calculator, and Planck's constant in every unit is on its own page.
Questions students ask
What is the de Broglie wavelength formula?
λ = h/p = h/mv, where h = 6.626 × 10⁻³⁴ J s. For a particle with kinetic energy K, λ = h/√(2mK).
What is the de Broglie wavelength of an electron accelerated through 100 V?
λ = 12.27/√100 = 1.227 Å, about 0.123 nm.
How does λ depend on the accelerating voltage?
λ ∝ 1/√V. Four times the voltage halves the wavelength.
Which has a larger de Broglie wavelength, an electron or a proton with the same kinetic energy?
The electron. λ = h/√(2mK), so the lighter particle has the longer wavelength: about √1836 ≈ 43 times longer.