de Broglie waves and uncertainty principle

Momentum $p$ is the central quantity: it is momentum, not speed or mass alone, that sets the wave behaviour of a particle. de Broglie postulated that a particle of momentum $p$ is described by a wave of wavelength

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

where $h = 6.626\times10^{-34}$ J·s is the Planck constant. Because, by the impulse law $\Delta \vec p = \vec F\,\Delta t$ (from mechanics), a force changes the momentum vector, any interaction that changes the magnitude $|p|$ also changes $\lambda$: accelerating a particle along its motion shortens its wave. A force perpendicular to the momentum changes only its direction, not $|p|$, so $\lambda = h/|p|$ stays the same. For macroscopic bodies the momentum is large and $\lambda$ is negligible (of order $10^{-34}$ m), whereas an electron with $v \sim 10^{6}$ m/s gives $\lambda \sim 0.7$ nm — comparable to atomic spacings, so an electron beam diffracts from a crystal. The de Broglie wave is a probability-amplitude wave, not a vibration of a medium: its intensity at a point is proportional to the probability of finding the particle there. A quantum state cannot have a sharp position and a sharp momentum at the same time: $\Delta x\,\Delta p \ge \hbar/2$, where $\hbar = h/2\pi$. This is not a measurement error but a property of the state itself: confining a particle to a region $\Delta x$ forces a non-zero spread of momenta, hence a non-zero kinetic energy — which is why atoms have a finite size and are stable.

An electron moves with speed $v = 1.0\times10^{6}$ m/s. The electron mass is $m_e = 9.11\times10^{-31}$ kg. Find the de Broglie wavelength of the electron.

An electron is accelerated from rest through a potential difference $U = 100$ V. Find its de Broglie wavelength and compare it with the interatomic spacing in a crystal ($d \approx 0.2$ nm). Electron charge $e = 1.60\times10^{-19}$ C, mass $m_e = 9.11\times10^{-31}$ kg.

An electron is confined to a region of size $\Delta x = 0.10$ nm (the order of an atomic size). Using the uncertainty relation $\Delta x\,\Delta p \ge \hbar/2$, estimate the minimum momentum spread and the minimum kinetic energy of the electron. Compare the result with the hydrogen ionization energy $13.6$ eV.

Goal: see how $\lambda = h/p$ spans dozens of orders of magnitude and how the uncertainty relation works.

Safety: calculations and an on-screen model only; no radiation sources or high voltage are needed. In particular, neither an electron source nor an accelerating voltage is required — all data come from the problem statement.

Part 1. Spreadsheet table. Create columns $m$, $v$, $p = mv$, $\lambda = h/p$. Fill in the rows:

• electron, $m_e = 9.11\times10^{-31}$ kg, $v = 1.0\times10^{6}$ m/s;
• the same electron at $v = 1.0\times10^{7}$ m/s (compare $\lambda$);
• proton, $m_p = 1.67\times10^{-27}$ kg, $v = 1.0\times10^{6}$ m/s;
• baseball, $m = 0.145$ kg, $v = 40$ m/s.

Part 2. Graph. Plot $\lg\lambda$ against $\lg p$ — from $\lambda = h/p$ this should be a straight line of slope $-1$. Check that all four points fall on one line.

Part 3. Confinement. For $\Delta x = 1.0$ nm, $0.10$ nm, and $0.010$ nm, compute $\Delta p \ge \hbar/(2\Delta x)$ and $K_{min} \approx (\Delta p)^2/(2m_e)$. Plot $K_{min}$ versus $\Delta x$ and state the conclusion: the tighter the region, the larger the energy.

Part 4 (optional). Open an interactive quantum-wave interference model (for example, an electron double-slit simulation) and check that shrinking the de Broglie wavelength brings the fringes on the screen closer together.

Show two ideas at once: (1) the de Broglie wavelength is inversely proportional to momentum and differs by many orders of magnitude between microscopic and macroscopic objects; (2) the more tightly a particle is localized, the wider its spread of momenta.

Two panels.

Left — a log–log graph: horizontal axis $p$ (from $10^{-26}$ to $10^{1}$ kg·m/s), vertical axis $\lambda = h/p$ (from $10^{-34}$ to $10^{-9}$ m). A straight line of slope $-1$ carries the marked points "electron at $10^{6}$ m/s", "electron at $100$ eV", "proton", "baseball"; a horizontal dashed line at $0.2$ nm is labelled "interatomic spacing $d \approx 0.2$ nm" — wave effects in a crystal are observable when $\lambda \approx d$ (for a first-order diffraction maximum $\lambda \le 2d$), not for every wavelength above that line. Points near the line (electron at $100$ eV: $\lambda \approx 0.12$ nm) correspond to observable diffraction; the "baseball" point far below the line shows no observable wave effects.

Right — a sketch: a wave packet (several sine periods under a Gaussian envelope) of width $\Delta x$; below it a horizontal double-headed arrow labelled "$\Delta x$". Beside it a fan of momentum arrows: a narrow envelope gives a wide fan (large $\Delta p$), a wide envelope gives a narrow fan; the caption reads $\Delta x\,\Delta p \ge \hbar/2$.

A football of mass $m = 0.43$ kg leaves a boot at $v = 25$ m/s, so its momentum is $p = mv = 0.43 \times 25 = 10.8$ kg·m/s. Its de Broglie wavelength is $\lambda = h/p = 6.63\times10^{-34}/10.8 \approx 6\times10^{-35}$ m. A proton is about $1.7\times10^{-15}$ m across, so the ball's matter wave is roughly $10^{20}$ times shorter than a single proton.

This is the de Broglie relation $\lambda = h/p$ in its simplest form: wavelength shrinks as momentum grows. Diffraction and interference appear only when $\lambda$ is comparable to the size of the obstacle or slit; for the ball $\lambda$ is $10^{20}$ times smaller than any gap it could pass through, so its wave nature is unobservable and Newtonian mechanics is enough.

In an electron microscope an electron is accelerated through $U = 100$ V and gains kinetic energy $E_k = eU = 100$ eV $= 1.60\times10^{-17}$ J. From $E_k = p^2/2m$ its momentum is $p = \sqrt{2 m_e E_k} = \sqrt{2 \times 9.11\times10^{-31} \times 1.60\times10^{-17}} \approx 5.4\times10^{-24}$ kg·m/s, so $\lambda = h/p \approx 6.63\times10^{-34}/5.4\times10^{-24} \approx 1.2\times10^{-10}$ m $= 0.12$ nm. That is exactly the spacing between atoms in a crystal, so a thin sample diffracts such a beam.

The de Broglie relation $\lambda = h/p$ together with the kinetic-energy link $E_k = p^2/2m$ converts an accelerating voltage into a wavelength. The resolving power is set by that wavelength: a $0.12$ nm wave reveals details finer than the roughly $550$ nm green light used in a light microscope.

In the silicon of a phone chip an electron is bound inside an atom about $1\times10^{-10}$ m across, so the uncertainty in its position is at most $\Delta x \approx 1\times10^{-10}$ m. The uncertainty principle demands $\Delta p \ge \hbar/(2\Delta x) = 1.055\times10^{-34}/(2\times10^{-10}) \approx 5.3\times10^{-25}$ kg·m/s. Even the calmest electron allowed by that momentum carries kinetic energy $\Delta p^2/(2m_e) \approx (5.3\times10^{-25})^2/(2\times9.11\times10^{-31}) \approx 1.5\times10^{-19}$ J $\approx 1$ eV.

This is the position–momentum form of the uncertainty principle $\Delta x\,\Delta p \ge \hbar/2$. Squeezing the electron into a smaller $\Delta x$ forces a larger $\Delta p$ and therefore more kinetic energy, so an electron cannot lie still on top of the nucleus. The resulting energy is of the same order as the few-electron-volt energies of chemical bonds, and the balance between this kinetic energy and the pull of the nucleus is what gives atoms their $\sim1\times10^{-10}$ m size.