Study Guide

de Broglie wavelength

CIE A-Level PhysicsΒ· Unit 26: Quantum PhysicsΒ· 15 min read

1. de Broglie Hypothesis and the de Broglie Equationβ˜…β˜…β˜†β˜†β˜†β± 5 min

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.

πŸ“˜ Definition

de Broglie wavelength

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

Example:

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

Ξ»=hp=hmv\lambda = \frac{h}{p} = \frac{h}{mv}
πŸ“ Worked Example

Calculate the de Broglie wavelength of an electron moving at . Use and .

  1. 1
    1. First calculate the electron's momentum:
  2. 2
    p=mev=(9.11Γ—10βˆ’31)(2.0Γ—106)=1.822Γ—10βˆ’24 kg m sβˆ’1p = m_e v = (9.11 \times 10^{-31})(2.0 \times 10^6) = 1.822 \times 10^{-24} \text{ kg m s}^{-1}
  3. 3
    1. Substitute into the de Broglie equation:
  4. 4
    Ξ»=hp=6.63Γ—10βˆ’341.822Γ—10βˆ’24=3.6Γ—10βˆ’10 m\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{1.822 \times 10^{-24}} = 3.6 \times 10^{-10} \text{ m}

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.

2. de Broglie Wavelength for Accelerated Charged Particlesβ˜…β˜…β˜…β˜†β˜†β± 7 min

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.

πŸ”¬ Derivation
Goal:

Derive the de Broglie wavelength formula for an electron accelerated through potential difference

Starting from:

Work done by electric field = kinetic energy gained by electron

  1. 1
    1. Equate work done to kinetic energy:
  2. 2
    eV=p22meeV = \frac{p^2}{2m_e}
  3. 3
    1. Rearrange to solve for momentum:
  4. 4
    p=2meeVp = \sqrt{2 m_e e V}
  5. 5
    1. Substitute into the general de Broglie equation:
  6. 6
    Ξ»=h2meeV\lambda = \frac{h}{\sqrt{2 m_e e V}}
Result:

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

πŸ“ Worked Example

Calculate the de Broglie wavelength of an electron accelerated through 150 V. Use , , .

  1. 1
    1. Substitute values into the derived formula:
  2. 2
    Ξ»=6.63Γ—10βˆ’342Γ—9.11Γ—10βˆ’31Γ—1.60Γ—10βˆ’19Γ—150\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.60 \times 10^{-19} \times 150}}
  3. 3
    1. Calculate the denominator first:
  4. 4
    denominator=4.37Γ—10βˆ’47=6.61Γ—10βˆ’24\text{denominator} = \sqrt{4.37 \times 10^{-47}} = 6.61 \times 10^{-24}
  5. 5
    1. Solve for wavelength:
  6. 6
    Ξ»=6.63Γ—10βˆ’346.61Γ—10βˆ’24β‰ˆ1.0Γ—10βˆ’10 m=0.1 nm\lambda = \frac{6.63 \times 10^{-34}}{6.61 \times 10^{-24}} \approx 1.0 \times 10^{-10} \text{ m} = 0.1 \text{ nm}

3. Experimental Confirmation: Electron Diffractionβ˜…β˜…β˜…β˜†β˜†β± 6 min

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.

πŸ“˜ Definition

Electron diffraction

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

πŸ“ Worked Example

A neutron beam is used to study atomic spacing in a crystal with inter-atomic spacing of . Estimate the speed of neutrons needed for maximum diffraction. Mass of neutron .

  1. 1
    1. Maximum diffraction occurs when
  2. 2
    1. Rearrange the de Broglie equation for speed:
  3. 3
    v=hmnΞ»v = \frac{h}{m_n \lambda}
  4. 4
    1. Substitute values and calculate:
  5. 5
    v=6.63Γ—10βˆ’34(1.67Γ—10βˆ’27)(2.8Γ—10βˆ’10)β‰ˆ1.4Γ—103 m sβˆ’1v = \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.

4. Common Pitfalls

Wrong move:

Using to calculate the wavelength of a particle

Why:

only applies to electromagnetic radiation, not matter waves

Correct move:

Always use for the wavelength of any moving particle

Wrong move:

Keeping accelerating potential in kV when substituting into the formula

Why:

All SI unit calculations require potential difference in volts, leading to a wavelength 1000 times smaller than the correct value

Correct move:

Always convert kV to V by multiplying by 1000 before substituting

Wrong move:

Claiming macroscopic objects do not have a de Broglie wavelength

Why:

All moving matter has a de Broglie wavelength, regardless of size

Correct move:

Explain that large mass gives large momentum, resulting in a wavelength too small to produce observable diffraction effects

Wrong move:

Using non-relativistic for very high energy (MeV) electrons

Why:

At speeds close to , relativistic effects make the non-relativistic kinetic energy approximation invalid

Correct move:

CIE 9702 almost always expects non-relativistic calculations; use the formula above unless explicitly told otherwise

5. Quick Reference Cheatsheet

Concept

Formula

Key Notes

General de Broglie wavelength

All moving particles

Accelerated electron

in volts, in metres

Maximum diffraction condition

Required for observable diffraction

Experimental proof

Electron diffraction through crystals

Confirms matter wave hypothesis

6. Frequently Asked

Is the de Broglie formula provided in the CIE data booklet?

No, you are required to remember the formula for exams.

Why can't we observe de Broglie wavelength for macroscopic objects?

Large mass gives very large momentum, so the wavelength is many orders of magnitude smaller than any detectable aperture, so no observable wave effects occur.

When this came up on past exams

AI-estimated based on syllabus patterns β€” cross-check with official past papers for accuracy. Use only as revision-focus signals.

  • 2022 Β· 2

    Calculate wavelength of accelerated electron

  • 2021 Β· 4

    Explain diffraction evidence for de Broglie

  • 2023 Β· 1

    MCQ wavelength calculation

Going deeper

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.