Dual nature of matter

Light already showed both faces. Planck's quantum theory and Einstein's explanation of the photoelectric effect had established that light — normally described by wave optics, with wavelength and frequency — also carries energy in discrete particle-like packets called photons, each with energy $E = h\nu$ and momentum $p = h\nu/c = h/\lambda$. Light was proven to be both wave and particle.

De Broglie reversed the logic. If a "wave" (light) can behave like a particle, de Broglie reasoned that a "particle" (matter) should equally be able to behave like a wave. For a particle of mass $m$ moving with velocity $v$, so that its momentum is $p = mv$, he proposed an associated wavelength

where: $\lambda$ is the de Broglie wavelength, $h = 6.626\times10^{-34}\ \text{J s}$ is Planck's constant, $m$ is the particle's mass, $v$ is its speed, and $p = mv$ is its momentum. The relation says wave character is inversely tied to momentum — heavy, fast objects get a vanishingly small $\lambda$, while light, slow-moving particles like electrons get a $\lambda$ large enough to matter.

Why we notice it for electrons but not cricket balls. A thrown ball has a mass around $0.15\ \text{kg}$, so even at modest speed its $\lambda$ works out to roughly $10^{-33}\ \text{m}$ — far smaller than any detector could ever resolve, so its wave nature is real but permanently unobservable. An electron, with mass $9.1\times10^{-31}\ \text{kg}$ moving at typical atomic speeds, has $\lambda$ of the order of an angstrom — comparable to atomic spacing — which is why beams of electrons diffract off crystals exactly as X-rays do, giving direct experimental proof of matter waves.

Heisenberg's uncertainty principle. Werner Heisenberg showed, independently of any instrument's precision, that the uncertainties in a particle's position ($\Delta x$) and momentum ($\Delta p$) must always satisfy

where: $\Delta x$ is the uncertainty in position, $\Delta p = m\,\Delta v$ is the uncertainty in momentum, and $h/4\pi$ is a fixed, non-zero lower bound. Pinning down position more tightly (smaller $\Delta x$) forces momentum to become correspondingly less certain (larger $\Delta p$), and vice versa — this is a property of nature at microscopic scale, not a measurement flaw.

Why Bohr's orbits could not survive. Bohr's model requires an electron with a precisely known position and a precisely known velocity at every instant, tracing a definite circular path. De Broglie's relation shows the electron has wave character, and a wave is inherently spread out, not located at one point; Heisenberg's principle shows position and momentum cannot both be pinned down even in principle. A sharply defined orbit is therefore not a physically meaningful concept for an electron — the atom must instead be described in terms of the probability of finding the electron in a given region, which is exactly the language of the quantum mechanical model and its orbitals.

Question. Calculate the de Broglie wavelength of an electron ($m = 9.1\times10^{-31}\ \text{kg}$) moving with a velocity of $2.2\times10^6\ \text{m s}^{-1}$ (a typical speed for an electron in a hydrogen atom's first Bohr orbit).

Step 1. Write the de Broglie relation: $\lambda = \dfrac{h}{mv}$.

Step 2. Compute the momentum: $mv = 9.1\times10^{-31} \times 2.2\times10^{6} = 2.00\times10^{-24}\ \text{kg m s}^{-1}$.

Step 3. Divide Planck's constant by this momentum: $\lambda = \dfrac{6.626\times10^{-34}}{2.00\times10^{-24}} = 3.31\times10^{-10}\ \text{m}$.

Answer: $\lambda = 3.31\times10^{-10}\ \text{m} = 3.31\ \text{Å}$ — comparable to atomic dimensions, which is exactly why electrons show observable wave behaviour inside atoms.

Competitive stretch

Question. Using the same electron, find the minimum uncertainty in its velocity if its position is known to within $\Delta x = 1.0\times10^{-10}\ \text{m}$ (the size of an atom).

Step 1. Write Heisenberg's principle with $\Delta p = m\,\Delta v$: $\Delta x \cdot m\,\Delta v \geq h/4\pi$, so $\Delta v \geq \dfrac{h}{4\pi\, m\, \Delta x}$.

Step 2. Substitute: denominator $= 4\pi \times 9.1\times10^{-31} \times 1.0\times10^{-10} = 1.144\times10^{-39}$.

Answer: $\Delta v \geq \dfrac{6.626\times10^{-34}}{1.144\times10^{-39}} \approx 5.8\times10^{5}\ \text{m s}^{-1}$ — an uncertainty comparable to the orbital speed itself, showing why a sharp orbit is meaningless once the electron's position is confined to atomic size.

• Assuming macroscopic objects "should" show observable wave behaviour because $\lambda = h/mv$ never gives exactly zero — the wavelength is real but astronomically small for large masses, hence undetectable in practice, not literally absent.
• Treating Heisenberg's uncertainty principle as a limitation of measuring instruments rather than a fundamental property of microscopic nature — better apparatus can never beat the $h/4\pi$ limit.
• Underestimating how completely the uncertainty principle breaks Bohr's model — it does not merely refine the fixed-orbit picture, it removes the very idea of a definite trajectory for the electron.
• Mixing up which two quantities the uncertainty relation links for the same particle at the same instant — it is $\Delta x$ and $\Delta p$ (or $\Delta v$) of one electron simultaneously, not two different particles or two different times.

Ask the tutor: want to see, side by side, why the numbers make electron waves observable but ball waves impossible — ask the AI tutor to compare the two de Broglie wavelength calculations.