How Do You Calculate the de Broglie Wavelength (λ = h/p)?

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Reasoning and Method

The de Broglie relationship states that every moving particle has an associated wavelength:

Here, $\lambda$ is the wavelength in metres, $h$ is Planck's constant, and $p$ is the particle's linear momentum in $\text{kg m s}^{-1}$. This is part of IB Physics E.2 Quantum physics, which is HL only.

Dimensional analysis confirms the unit: $h$ has units $\text{J s} = \text{kg m}^2\text{s}^{-1}$. Dividing by $\text{kg m s}^{-1}$ leaves metres. Before calculating, convert grams to kilograms, electronvolts to joules when using classical kinetic energy, and nanometres to metres if required.

Choose the momentum expression that matches the information given:

If kinetic energy is supplied, combine $E_k = \frac{1}{2}mv^2$ with $p = mv$. This gives $p = \sqrt{2mE_k}$, so $\lambda = \frac{h}{\sqrt{2mE_k}}$. This route applies when the particle is non-relativistic.

For example, calculate the wavelength of an electron moving at $2.00 \times 10^6\ \text{m s}^{-1}$. Using $m_e = 9.110 \times 10^{-31}\ \text{kg}$:

Therefore,

This is $0.364\ \text{nm}$. Its atomic-scale size explains why electron diffraction can reveal crystal structure and provide evidence for the wave nature of matter.

The relationship also predicts that increasing momentum decreases wavelength. If momentum doubles, wavelength halves. Observable diffraction occurs when the wavelength is comparable to the spacing in the obstacle or crystal, which is why electrons show diffraction but everyday objects do not.

A common misconception is that de Broglie wavelength depends only on speed. It depends on momentum, so both mass and speed matter. At the same speed, a more massive particle has a shorter wavelength.

Exam Technique

For a Calculate question, show the momentum calculation, substitution into $\lambda = h/p$, and final wavelength with a unit. Use the data booklet value of $h$, convert all quantities to SI units, and do not confuse momentum $p$ with kinetic energy.