Wave Nature of Matter — de Broglie Hypothesis

The de Broglie hypothesis

By the early twentieth century the dual nature of light was firmly established. Interference, diffraction and polarisation revealed its wave character, while the photoelectric effect and the Compton effect showed that radiation could behave as a stream of particles called photons. A natural question followed: if radiation is dual, might the particles of nature — electrons, protons and the rest — also exhibit wave-like behaviour?

In 1924 the French physicist Louis Victor de Broglie answered in the affirmative. He reasoned that nature is symmetrical, and that the two fundamental physical entities, matter and energy, should share that symmetry. If radiation shows a dual aspect, so should matter. He therefore proposed that a moving particle of matter has a wave associated with it, now called a matter wave or de Broglie wave.

The hypothesis was bold because it assigned a wavelength to objects that classical physics treated purely as localised particles. As NIOS §25.4 records, de Broglie's doctoral thesis containing this idea was initially met with scepticism, yet within a few years experiment confirmed it, and in 1929 he was awarded the Nobel Prize in Physics for the discovery of the wave nature of electrons.

The de Broglie wavelength

De Broglie proposed that the wavelength $\lambda$ associated with a particle of momentum $p$ is

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

where $h$ is Planck's constant ($6.63 \times 10^{-34}$ J s), $m$ is the mass of the particle and $v$ its speed. This is the de Broglie relation. Its elegance lies in the way it bridges two pictures: the left-hand side, $\lambda$, is a wave attribute, while the right-hand side carries $p$, the characteristic attribute of a particle. Planck's constant is the bridge that links them.

The relation is inverse: a larger momentum gives a smaller wavelength. For a fixed particle, that means a heavier or faster object has a shorter matter wavelength. Plotted against momentum, $\lambda$ traces a rectangular hyperbola.

The de Broglie wavelength does not depend on the charge or the chemical nature of the particle. Two particles of the same momentum — say an electron and a proton arranged to have equal $p$ — have the same $\lambda$. Charge matters only indirectly, through how it is used to set up the particle's momentum, as the next sections make clear.

Wavelength from kinetic energy

In most problems we know a particle's kinetic energy rather than its speed directly. For a non-relativistic particle of mass $m$ and kinetic energy $K$,

$$K = \tfrac{1}{2}mv^2 = \frac{p^2}{2m} \quad\Longrightarrow\quad p = \sqrt{2mK}$$

Substituting into the de Broglie relation gives the energy form of the wavelength:

$$\lambda = \frac{h}{\sqrt{2mK}}$$

This is the most versatile form for NEET. For a fixed mass, $\lambda \propto 1/\sqrt{K}$, so a fourfold rise in kinetic energy halves the wavelength. The relation also explains why neutrons in thermal equilibrium, where $K = \tfrac{3}{2}kT$, acquire a temperature-dependent wavelength $\lambda = h/\sqrt{3mkT}$ — a result tested verbatim in NEET 2017.

The accelerated electron

The most common laboratory way to control a particle's wavelength is to accelerate a charged particle through a known potential difference. An electron of charge $q = e$ accelerated from rest through a potential difference $V$ gains kinetic energy $K = eV$. Its momentum is therefore $p = \sqrt{2meV}$, and

$$\lambda = \frac{h}{\sqrt{2mqV}}$$

Substituting the constants for the electron ($h = 6.63 \times 10^{-34}$ J s, $m = 9.11 \times 10^{-31}$ kg, $e = 1.602 \times 10^{-19}$ C) collapses the formula into a single working expression:

$$\lambda = \frac{1.227}{\sqrt{V}}\ \text{nm} \qquad (V \text{ in volts})$$

For $V = 100$ V this gives $\lambda \approx 0.123$ nm, comparable to X-ray wavelengths and to the spacing of atomic planes in a crystal. That coincidence is precisely why a beam of accelerated electrons can be diffracted by a crystal lattice — the experimental confirmation taken up in the Davisson–Germer experiment.

Why macroscopic objects show no wave

The de Broglie relation applies to every moving body, including a cricket ball or a bullet. The reason we never observe their waves is the smallness of Planck's constant. For an ordinary object the momentum $mv$ is enormous compared with $h$, so $\lambda = h/mv$ becomes vanishingly small — far below any length scale we can probe.

NCERT makes this concrete with a ball of mass $0.12$ kg moving at $20$ m/s, giving $p = 2.40$ kg m/s and $\lambda = 2.76 \times 10^{-34}$ m. NIOS notes the same for a $50$ g ball rolling at $0.2$ m/s, whose wavelength is about $6.6 \times 10^{-32}$ m. These are immeasurable. Only in the sub-atomic domain, where masses and hence momenta are tiny, does $\lambda$ rise to the order of atomic-plane spacings and become significant.

Consistency with the photon

Equation $\lambda = h/p$ is a hypothesis for matter, testable only by experiment. It gains credibility, however, from the fact that it is automatically obeyed by a photon. For a photon the momentum is $p = h\nu/c$, so

$$\lambda = \frac{h}{p} = \frac{h}{h\nu/c} = \frac{c}{\nu}$$

which is exactly the wavelength of the electromagnetic radiation of which the photon is a quantum. The de Broglie wavelength of a photon thus equals its electromagnetic wavelength. A single relation, $\lambda = h/p$, therefore unifies the wavelength of light and the wavelength of matter — the symmetry de Broglie set out to capture.