de-Broglie Wavelength

If light can behave like a particle, can matter behave like a wave? Louis de Broglie (1924) proposed that all moving particles have an associated wavelength — this is the foundation of matter waves.

The Hypothesis

Every moving particle has a wavelength given by:

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

Where:

Symbol Meaning
$\lambda$ de-Broglie wavelength (m)
$h$ Planck's constant ($6.63 \times 10^{-34}$ J·s)
$p$ momentum (kg·m/s)
$m$ mass (kg)
$v$ velocity (m/s)

From Kinetic Energy

When velocity is not given directly:

$$\lambda = \frac{h}{\sqrt{2m(KE)}}$$

Derivation

From $KE = \frac{1}{2}mv^2$: $$v = \sqrt{\frac{2(KE)}{m}}$$

Substituting into $\lambda = \frac{h}{mv}$: $$\lambda = \frac{h}{m\sqrt{\frac{2(KE)}{m}}} = \frac{h}{\sqrt{2m(KE)}}$$

Key Implications

Small vs Large Objects

Object Mass Velocity de-Broglie Wavelength
Electron $9.11 \times 10^{-31}$ kg $10^6$ m/s $\sim 10^{-10}$ m (atomic scale)
Baseball 0.145 kg 40 m/s $\sim 10^{-34}$ m (undetectable)
Human 70 kg 1 m/s $\sim 10^{-35}$ m (undetectable)

Why we don't see humans as waves: The wavelength is astronomically small for macroscopic objects.

Wave Nature is Observable For

  • Electrons (electron microscopes)
  • Atoms (Bose-Einstein condensates)
  • Molecules (interference experiments)

Relationship with Energy

  • Higher KE → Higher momentum → Shorter wavelength
  • Fast-moving particles behave less like waves

Experimental Verification

Electron Diffraction (1927)

Davisson-Germer experiment: electrons showed diffraction patterns — proof of wave nature.

Electron Microscope

Uses electron waves (much shorter than light) to achieve higher resolution.

Example Calculations

Example 1: From Velocity

Find the de-Broglie wavelength of an electron moving at $v = 2.0 \times 10^6$ m/s.

$$\lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{(9.11 \times 10^{-31})(2.0 \times 10^6)} = 3.64 \times 10^{-10} \text{ m}$$

Example 2: From Kinetic Energy

Find $\lambda$ for an electron with $KE = 150$ eV.

Step 1: Convert to joules $$KE = 150 \times 1.602 \times 10^{-19} = 2.403 \times 10^{-17} \text{ J}$$

Step 2: Calculate $$\lambda = \frac{h}{\sqrt{2m(KE)}} = \frac{6.63 \times 10^{-34}}{\sqrt{2(9.11 \times 10^{-31})(2.403 \times 10^{-17})}} \approx 1.0 \times 10^{-10} \text{ m}$$

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