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CU-101 / L11 · lecture

Wave–Particle Duality and de Broglie

~5 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with the Bohr model’s quantisation of angular momentum and recall that light exhibits both wave-like interference and particle-like photoelectric effects.

Why this matters

The de Broglie hypothesis bridges the gap between classical particles and quantum waves, explaining why electrons are confined to discrete orbits.

In the previous lecture, we noted that the Bohr model worked for hydrogen but lacked a fundamental justification for why angular momentum must be quantised. Louis de Broglie proposed in 1924 that if light (traditionally a wave) has particle properties, then matter (traditionally particles) must have wave properties. This insight, known as wave–particle duality, provides the physical reason for quantisation: only standing waves are stable. Without this concept, the periodic table’s structure and the stability of atoms remain unexplained mysteries rather than logical consequences of wave mechanics.

The Core Idea: Matter as Waves

Imagine a guitar string. It can vibrate, but only at specific frequencies that create standing waves—patterns where the ends remain fixed. If you try to create a wave that does not fit an integer number of half-wavelengths between the fixed ends, the wave interferes destructively with itself and disappears. De Broglie suggested that electrons orbiting a nucleus behave like such a string. For an electron orbit to be stable, the circumference of the orbit must contain a whole number of electron wavelengths. If the wavelength does not fit perfectly, the electron wave cancels itself out, and that orbit is forbidden. This simple geometric requirement naturally leads to the quantisation of energy levels, providing the physical basis for the Bohr model’s arbitrary postulate.

The de Broglie Wavelength

De Broglie proposed that any particle with momentum \( p \) has an associated wavelength \( \lambda \), given by the relation:

\[ \lambda = \frac{h}{p} \]

where \( h \) is Planck’s constant (\( 6.626 \times 10^{-34}\ \mathrm{J\,s} \)) and \( p \) is the momentum of the particle. For a non-relativistic particle of mass \( m \) moving at velocity \( v \), the momentum is \( p = mv \), so the wavelength can be written as:

\[ \lambda = \frac{h}{mv} \]

This equation reveals a crucial inverse relationship: the faster and heavier a particle is, the shorter its wavelength. For macroscopic objects, such as a baseball, the mass is so large that the wavelength is infinitesimally small, far smaller than any atomic scale, which is why we do not observe wave-like behaviour in everyday life. However, for electrons, which have a very small mass (\( 9.109 \times 10^{-31}\ \mathrm{kg} \)), the wavelength can be comparable to the size of atoms, making quantum effects observable.

Standing Waves and Quantisation

To see how this leads to quantisation, consider an electron in a circular orbit of radius \( r \). The condition for a stable standing wave is that the circumference of the orbit must be an integer multiple of the wavelength:

\[ 2\pi r = n\lambda \]

where \( n \) is a positive integer (\( n = 1, 2, 3, \dots \)). Substituting the de Broglie expression \( \lambda = h/p \) into this condition gives:

\[ 2\pi r = n \frac{h}{p} \]

Rearranging this equation yields:

\[ pr = \frac{nh}{2\pi} \]

Since \( p = mv \), the product \( pr \) is the angular momentum \( L \) of the electron. Thus, we obtain:

\[ L = \frac{nh}{2\pi} \]

or, using \( \hbar = h/2\pi \):

\[ L = n\hbar \]

This is precisely the quantisation condition postulated by Bohr. The de Broglie hypothesis therefore provides a physical justification for Bohr’s first postulate: angular momentum is quantised because only orbits that support standing electron waves are stable.

Experimental Verification

The wave nature of electrons was confirmed experimentally by Clinton Davisson and Lester Germer in 1927. They fired a beam of electrons at a nickel crystal and observed an interference pattern, analogous to X-ray diffraction. The angles at which the intensity of the reflected electrons peaked matched the predictions of Bragg’s law, which is used for X-rays. This demonstrated that electrons behave as waves with wavelengths given by \( \lambda = h/p \). Later, the double-slit experiment with electrons showed that even single electrons, sent one at a time, build up an interference pattern over time, confirming that each individual electron exhibits wave-like behaviour.

Tier 3 · watch out

A common misconception is that the de Broglie wavelength represents a physical oscillation of the electron’s charge density in space, like a water wave. In reality, the wave function is a probability amplitude; the wavelength determines the spatial frequency of the probability distribution, not a mechanical vibration of the particle itself. The electron is not "smeared out" as a fluid, but its detection probability exhibits wave-like interference. C

Worked example

Problem. Calculate the de Broglie wavelength of an electron with a mass of \( 9.109 \times 10^{-31}\ \mathrm{kg} \) moving at a velocity of \( 2.00 \times 10^6\ \mathrm{m\,s^{-1}} \). Use \( h = 6.626 \times 10^{-34}\ \mathrm{J\,s} \).

  1. Identify the given values: \( m = 9.109 \times 10^{-31}\ \mathrm{kg} \), \( v = 2.00 \times 10^6\ \mathrm{m\,s^{-1}} \), \( h = 6.626 \times 10^{-34}\ \mathrm{J\,s} \).
  2. Calculate the momentum \( p = mv \): \( p = (9.109 \times 10^{-31}\ \mathrm{kg}) \times (2.00 \times 10^6\ \mathrm{m\,s^{-1}}) = 1.8218 \times 10^{-24}\ \mathrm{kg\,m\,s^{-1}} \).
  3. Apply the de Broglie equation \( \lambda = h/p \): \( \lambda = \frac{6.626 \times 10^{-34}\ \mathrm{J\,s}}{1.8218 \times 10^{-24}\ \mathrm{kg\,m\,s^{-1}}} \).
  4. Perform the division: \( \lambda \approx 3.637 \times 10^{-10}\ \mathrm{m} \).
  5. Convert to picometres for context: \( 3.637 \times 10^{-10}\ \mathrm{m} = 363.7\ \mathrm{pm} \).
\[ \lambda = 3.64 \times 10^{-10}\ \mathrm{m} \]

Units check. The units of \( h \) are \( \mathrm{J\,s} = \mathrm{kg\,m^2\,s^{-2}\,s} = \mathrm{kg\,m^2\,s^{-1}} \). Dividing by momentum \( \mathrm{kg\,m\,s^{-1}} \) yields \( \mathrm{m} \), which is correct for a wavelength.

CHECK YOURSELF

Three quick tests

  1. How does the de Broglie wavelength change if the velocity of an electron is doubled?
    AnswerIt is halved, because \( \lambda \) is inversely proportional to \( v \).
  2. Why do we not observe wave-like behaviour for a tennis ball?
    AnswerIts mass is so large that its de Broglie wavelength is extremely small, far smaller than any detectable scale.
  3. What physical condition does the standing wave requirement impose on the electron orbit?
    AnswerThe circumference must be an integer multiple of the wavelength, leading to quantised angular momentum.
UNIT

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NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Results cited: — · Threads: quantum, structure