recall and use λ = h / p

Wave‑Particle Duality

Learning Objective

Recall and apply the de Broglie relation

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

where λ is the wavelength associated with a particle, h is Planck’s constant (6.626 × 10⁻³⁴ J·s) and p is the linear momentum of the particle.

Key Concepts

• All matter exhibits both wave‑like and particle‑like properties.
• The wave aspect is characterised by a wavelength given by the de Broglie equation.
• Momentum p is the product of mass m and velocity v: $p = mv$.
• For photons, $p = \frac{E}{c}$ and $E = hf$, leading to $\lambda = \frac{c}{f}$, which is consistent with the de Broglie formula.

Derivation of the de Broglie Wavelength

1. Start from the photon relations:
   • Energy: $E = hf$
   • Momentum: $p = \frac{E}{c}$
   • Wave‑speed: $c = f\lambda$
2. Combine them: $$p = \frac{hf}{c} = \frac{h}{\lambda}$$
3. Rearrange to obtain de Broglie’s hypothesis for any particle: $$\lambda = \frac{h}{p}$$

Applying the Formula

To find the wavelength of a particle, follow these steps:

1. Determine the particle’s mass m (kg) and speed v (m s⁻¹).
2. Calculate its momentum: $p = mv$.
3. Insert $p$ into the de Broglie equation to obtain $\lambda$.

Worked Example

Calculate the de Broglie wavelength of an electron accelerated through a potential difference of 150 V.

Solution:

1. Electron charge: $e = 1.602\times10^{-19}\,\text{C}$.
2. Kinetic energy gained: $E_k = eV = (1.602\times10^{-19})(150) = 2.40\times10^{-17}\,\text{J}$.
3. Relate kinetic energy to momentum (non‑relativistic): $$E_k = \frac{p^{2}}{2m} \;\Rightarrow\; p = \sqrt{2mE_k}$$ with $m_e = 9.11\times10^{-31}\,\text{kg}$.
4. Compute $p$: $$p = \sqrt{2(9.11\times10^{-31})(2.40\times10^{-17})} = 6.60\times10^{-24}\,\text{kg·m s}^{-1}$$
5. Finally, $$\lambda = \frac{h}{p} = \frac{6.626\times10^{-34}}{6.60\times10^{-24}} = 1.00\times10^{-10}\,\text{m}$$ (approximately 0.10 nm).

Comparison of Wavelengths for Different Particles

Particle	Mass (kg)	Speed (m s⁻¹)	Momentum $p$ (kg·m s⁻¹)	Wavelength $λ$ (m)
Electron (150 V)	9.11 × 10⁻³¹	2.3 × 10⁶	2.1 × 10⁻²⁴	3.2 × 10⁻¹⁰
Proton (1 MeV)	1.67 × 10⁻²⁷	1.38 × 10⁷	2.3 × 10⁻²⁰	2.9 × 10⁻¹⁴
Macroscopic ball (0.1 kg, 1 m s⁻¹)	1.0 × 10⁻¹	1	1.0 × 10⁻¹	6.6 × 10⁻³³

Experimental Evidence

• Electron diffraction through thin crystals demonstrates wave‑like interference patterns.
• Neutron interferometry shows similar behaviour for massive neutral particles.
• Large molecules (e.g., C₆₀ fullerene) have been diffracted, confirming that the de Broglie wavelength applies to complex systems.

Common Misconceptions

• “Only light is a wave.” – All particles have an associated wavelength; the magnitude determines whether wave effects are observable.
• “The wavelength is a physical size.” – It represents the spatial periodicity of the probability amplitude, not a literal length of the particle.
• “Higher speed always means larger wavelength.” – Since $p = mv$, increasing speed increases momentum, which actually *decreases* the wavelength.

Practice Questions

1. Calculate the de Broglie wavelength of a neutron moving at $2.0\times10^{5}\,\text{m s}^{-1}$. (Neutron mass $=1.675\times10^{-27}\,\text{kg}$.)
2. A beam of electrons with kinetic energy $100\,\text{eV}$ is incident on a double‑slit apparatus with slit separation $0.5\,\mu\text{m}$. Determine whether an interference pattern can be observed.
3. Explain why macroscopic objects do not exhibit observable diffraction, using the de Broglie relation.