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):

   \$Ek = \frac{p^{2}}{2m} \;\Rightarrow\; p = \sqrt{2mEk}\$

   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

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.