# de Broglie wavelength

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u26-de-broglie-wavelength/

This module explains de Broglie's hypothesis that all matter has wave-like properties, how to calculate the de Broglie wavelength of moving particles, and how electron diffraction provides experimental proof of matter waves.

**Prerequisites:** [Momentum and kinetic energy](https://www.owlsprep.com/study/cie-9702-u03-momentum-kinetic-energy/); [Wave diffraction](https://www.owlsprep.com/study/cie-9702-u08-wave-properties-diffraction/)

## Learning objectives

- State the de Broglie hypothesis of matter waves
- Calculate the de Broglie wavelength of moving particles
- Derive the wavelength formula for accelerated charged particles
- Explain how electron diffraction confirms de Broglie's hypothesis

## de Broglie Hypothesis and the de Broglie Equation

In 1924, Louis de Broglie proposed that wave-particle duality is not unique to electromagnetic radiation: all moving particles exhibit both particle and wave properties. These wave-like properties are described as *matter waves*.

**de Broglie wavelength** — The wavelength associated with the matter wave of a moving particle, directly related to the particle's momentum.

*Notation:* $\lambda$

*Example:* A 100 V accelerated electron has a de Broglie wavelength ~0.1 nm, matching atomic spacing in crystals.

$$\lambda = \frac{h}{p} = \frac{h}{mv}$$

**Worked example:** Calculate the de Broglie wavelength of an electron moving at $2.0 \times 10^6 \text{ m s}^{-1}$. Use $m_e = 9.11 \times 10^{-31} \text{ kg}$ and $h = 6.63 \times 10^{-34} \text{ J s}$.

1. 1. First calculate the electron's momentum:
2. $$p = m_e v = (9.11 \times 10^{-31})(2.0 \times 10^6) = 1.822 \times 10^{-24} \text{ kg m s}^{-1}$$
3. 2. Substitute into the de Broglie equation:
4. $$\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.822 \times 10^{-24}} = 3.6 \times 10^{-10} \text{ m}$$

> **tip**
>
> Always check units: momentum must be in $\text{kg m s}^{-1}$ and $h$ in J s to get wavelength in metres.

> **Exam tip:** You may be asked to state de Broglie's hypothesis in 2-3 marks: always mention that *all moving matter* has a wavelength related to its momentum.

## de Broglie Wavelength for Accelerated Charged Particles

The most common exam question asks you to calculate the de Broglie wavelength of an electron accelerated through a known potential difference. We can derive a simplified formula for this case using conservation of energy.

**Derivation:** Derive the de Broglie wavelength formula for an electron accelerated through potential difference $V$

*Starting from:* Work done by electric field = kinetic energy gained by electron

1. 1. Equate work done to kinetic energy:
2. $$eV = \frac{p^2}{2m_e}$$
3. 2. Rearrange to solve for momentum:
4. $$p = \sqrt{2 m_e e V}$$
5. 3. Substitute into the general de Broglie equation:
6. $$\lambda = \frac{h}{\sqrt{2 m_e e V}}$$

*Conclusion:* This formula lets you calculate wavelength directly from the accelerating potential, without calculating speed first.

**Worked example:** Calculate the de Broglie wavelength of an electron accelerated through 150 V. Use $e = 1.60 \times 10^{-19} \text{ C}$, $m_e = 9.11 \times 10^{-31} \text{ kg}$, $h = 6.63 \times 10^{-34} \text{ J s}$.

1. 1. Substitute values into the derived formula:
2. $$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 150}}$$
3. 2. Calculate the denominator first:
4. $$\text{denominator} = \sqrt{4.37 \times 10^{-47}} = 6.61 \times 10^{-24}$$
5. 3. Solve for wavelength:
6. $$\lambda = \frac{6.63 \times 10^{-34}}{6.61 \times 10^{-24}} \approx 1.0 \times 10^{-10} \text{ m} = 0.1 \text{ nm}$$

## Experimental Confirmation: Electron Diffraction

de Broglie's hypothesis was confirmed in 1927 by Davisson and Germer, who observed that electrons scattered off a crystalline nickel target produced a clear diffraction pattern. Diffraction is an exclusively wave property, so this proved that electrons exhibit wave-like behaviour.

Maximum diffraction occurs when the de Broglie wavelength of the incident particles is approximately equal to the size of the diffracting gap. For crystals, the gap is the inter-atomic spacing (~0.1 nm), which matches the wavelength of electrons accelerated through ~100 V, as we saw in the previous example.

**Electron diffraction** — The diffraction of electrons by a crystalline material, which provides direct experimental proof of de Broglie's matter wave hypothesis.

**Worked example:** A neutron beam is used to study atomic spacing in a crystal with inter-atomic spacing of $2.8 \times 10^{-10} \text{ m}$. Estimate the speed of neutrons needed for maximum diffraction. Mass of neutron $m_n = 1.67 \times 10^{-27} \text{ kg}$.

1. 1. Maximum diffraction occurs when $\lambda = \text{inter-atomic spacing} = 2.8 \times 10^{-10} \text{ m}$
2. 2. Rearrange the de Broglie equation for speed:
3. $$v = \frac{h}{m_n \lambda}$$
4. 3. Substitute values and calculate:
5. $$v = \frac{6.63 \times 10^{-34}}{(1.67 \times 10^{-27})(2.8 \times 10^{-10})} \approx 1.4 \times 10^3 \text{ m s}^{-1}$$

> **Exam tip:** When asked to explain why electron diffraction supports de Broglie's hypothesis, explicitly state that diffraction is a wave property, proving particles have wave behaviour.

## Common pitfalls

- **Wrong:** Using $\lambda = c/f$ to calculate the wavelength of a particle
  - Why it fails: $c = f\lambda$ only applies to electromagnetic radiation, not matter waves
  - Correct: Always use $\lambda = h/p$ for the wavelength of any moving particle
- **Wrong:** Keeping accelerating potential in kV when substituting into the formula
  - Why it fails: All SI unit calculations require potential difference in volts, leading to a wavelength 1000 times smaller than the correct value
  - Correct: Always convert kV to V by multiplying by 1000 before substituting
- **Wrong:** Claiming macroscopic objects do not have a de Broglie wavelength
  - Why it fails: All moving matter has a de Broglie wavelength, regardless of size
  - Correct: Explain that large mass gives large momentum, resulting in a wavelength too small to produce observable diffraction effects
- **Wrong:** Using non-relativistic $E_k = p^2/(2m)$ for very high energy (MeV) electrons
  - Why it fails: At speeds close to $c$, relativistic effects make the non-relativistic kinetic energy approximation invalid
  - Correct: CIE 9702 almost always expects non-relativistic calculations; use the formula above unless explicitly told otherwise

## Cheatsheet

| Concept | Formula | Key Notes |
| --- | --- | --- |
| General de Broglie wavelength | $\lambda = \frac{h}{mv}$ | All moving particles |
| Accelerated electron | $\lambda = \frac{h}{\sqrt{2 m_e e V}}$ | $V$ in volts, $\lambda$ in metres |
| Maximum diffraction condition | $\lambda \approx \text{gap spacing}$ | Required for observable diffraction |
| Experimental proof | Electron diffraction through crystals | Confirms matter wave hypothesis |

## What's next

de Broglie's hypothesis of matter waves forms the foundation of modern quantum physics. It confirmed wave-particle duality, paved the way for the development of quantum mechanics, and enabled revolutionary technologies such as transmission electron microscopes, which use the small de Broglie wavelength of high-energy electrons to resolve atomic-scale structures that are impossible to see with light microscopes. Understanding this concept is critical for exploring further quantum phenomena like the uncertainty principle and nuclear physics.

- [Wave-particle duality](https://www.owlsprep.com/study/cie-9702-u26-wave-particle-duality/)
- [Nuclear physics](https://www.owlsprep.com/study/cie-9702-u27-overview/)

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