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de Broglie Hypothesis: Wave Nature of Matter

The de Broglie hypothesis states that every moving material particle is associated with a matter wave whose wavelength is lambda = h/p, where p is the momentum of the particle.

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Student-friendly explanation

Light shows both wave and particle behaviour, and de Broglie proposed that moving particles should also show wave behaviour. For a non-relativistic particle of mass m moving with speed v, p = mv, so lambda = h/mv. The wavelength is noticeable for microscopic particles such as electrons but extremely small for everyday objects because their momentum is large. Electron diffraction provides evidence for the wave nature of matter.

How to write this in exams

  1. 1

    Start with the exact idea

    The de Broglie hypothesis states that every moving material particle is associated with a matter wave whose wavelength is lambda = h/p, where p is the momentum of the particle.

  2. 2

    Then show how to use it

    Identify the moving particle, choose p = mv or momentum from acceleration voltage, substitute SI values, calculate lambda, and compare the result with atomic dimensions if the question asks for physical significance.

  3. 3

    Add one concrete example

    An electron accelerated through a potential difference has a measurable de Broglie wavelength, which allows electron beams to produce diffraction patterns from crystals.

  4. 4

    Avoid this incomplete answer

    Wrong: A particle at rest has a de Broglie wavelength h/m. Correct: de Broglie wavelength is associated with momentum; if p = 0, lambda = h/p is not meaningful as a finite matter wavelength for motion.

Definition

The de Broglie hypothesis states that every moving material particle is associated with a matter wave whose wavelength is lambda = h/p, where p is the momentum of the particle.

Example

An electron accelerated through a potential difference has a measurable de Broglie wavelength, which allows electron beams to produce diffraction patterns from crystals.

Rule to remember

de Broglie wavelength: lambda = h/p. For a non-relativistic particle, lambda = h/mv. For a charged particle accelerated through potential V, kinetic energy qV = p^2/(2m), so lambda = h/sqrt(2mqV). For an electron, q = e. SI units: lambda in m, h in J s, p in kg m s^-1, m in kg, v in m s^-1, q in C, and V in volt. Use only when the particle is moving and non-relativistic conditions are assumed for mv form.

Memory hook

Matter wavelength becomes visible when momentum becomes tiny.

Examples and method

Worked example

An electron is accelerated through 100 V. Using lambda = h/sqrt(2meV), lambda = (6.63 x 10^-34)/sqrt(2 x 9.11 x 10^-31 x 1.6 x 10^-19 x 100). The denominator is about 5.40 x 10^-24 kg m s^-1, so lambda = 1.23 x 10^-10 m. This wavelength is of atomic size, so electron diffraction is possible.

Method to apply

Identify the moving particle, choose p = mv or momentum from acceleration voltage, substitute SI values, calculate lambda, and compare the result with atomic dimensions if the question asks for physical significance.

Diagram support

For Davisson-Germer evidence, a diagram may show an electron beam incident on a nickel crystal, scattered electrons, detector, and intensity variation with angle. For formula use, a wave drawn along the particle path can support the idea.

How CBSE asks it

Asked as a statement of hypothesis, derivation of lambda = h/sqrt(2meV), numerical wavelength of an electron, comparison with macroscopic objects, and significance of Davisson-Germer experiment.

Avoid common mistakes

Common confusion

Students often apply lambda = h/mv to photons. For photons, use p = h/lambda or E/c; lambda = h/p is general, but mv is not used for photons because photons have no rest mass.

Common wrong answer

Wrong: A particle at rest has a de Broglie wavelength h/m. Correct: de Broglie wavelength is associated with momentum; if p = 0, lambda = h/p is not meaningful as a finite matter wavelength for motion.

Exam tip

For electron acceleration numericals, first find momentum from kinetic energy or use the standard electron relation if allowed. Always keep mass in kg, charge in coulomb, potential in volt, and wavelength in metre.

Quick check

Why is de Broglie wavelength not observed for a moving cricket ball in ordinary conditions?

A cricket ball has very large momentum compared with an electron, so lambda = h/p gives an extremely small wavelength. The wave behaviour is therefore not observable in ordinary situations, while it becomes important for microscopic particles.

Answer writing and exam use

1-mark answer

The de Broglie hypothesis states that every moving material particle is associated with a matter wave whose wavelength is lambda = h/p, where p is the momentum of the particle.

2-mark answer

The de Broglie hypothesis states that every moving material particle is associated with a matter wave whose wavelength is lambda = h/p, where p is the momentum of the particle. de Broglie wavelength: lambda = h/p. For a non-relativistic particle, lambda = h/mv. For a charged particle accelerated through potential V, kinetic energy qV = p^2/(2m), so lambda = h/sqrt(2mqV). For an electron, q = e. SI units: lambda in m, h in J s, p in kg m s^-1, m in kg, v in m s^-1, q in C, and V in volt. Use only when the particle is moving and non-relativistic conditions are assumed for mv form. An electron accelerated through a potential difference has a measurable de Broglie wavelength, which allows electron beams to produce diffraction patterns from crystals.

3-mark answer

Light shows both wave and particle behaviour, and de Broglie proposed that moving particles should also show wave behaviour. For a non-relativistic particle of mass m moving with speed v, p = mv, so lambda = h/mv. The wavelength is noticeable for microscopic particles such as electrons but extremely small for everyday objects because their momentum is large. Electron diffraction provides evidence for the wave nature of matter. de Broglie wavelength: lambda = h/p. For a non-relativistic particle, lambda = h/mv. For a charged particle accelerated through potential V, kinetic energy qV = p^2/(2m), so lambda = h/sqrt(2mqV). For an electron, q = e. SI units: lambda in m, h in J s, p in kg m s^-1, m in kg, v in m s^-1, q in C, and V in volt. Use only when the particle is moving and non-relativistic conditions are assumed for mv form. An electron is accelerated through 100 V. Using lambda = h/sqrt(2meV), lambda = (6.63 x 10^-34)/sqrt(2 x 9.11 x 10^-31 x 1.6 x 10^-19 x 100). The denominator is about 5.40 x 10^-24 kg m s^-1, so lambda = 1.23 x 10^-10 m. This wavelength is of atomic size, so electron diffraction is possible. Asked as a statement of hypothesis, derivation of lambda = h/sqrt(2meV), numerical wavelength of an electron, comparison with macroscopic objects, and significance of Davisson-Germer experiment. Wrong: A particle at rest has a de Broglie wavelength h/m. Correct: de Broglie wavelength is associated with momentum; if p = 0, lambda = h/p is not meaningful as a finite matter wavelength for motion.
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