De Broglie matter waves: HSC Physics Module 8

Kinetic energy gained: $E_K = eV = 1.60 \times 10^{-19} \times 150 = 2.40 \times 10^{-17}$ J.

Speed (non-relativistic at this energy):

$v = \sqrt{2 E_K / m_e} = \sqrt{2 \times 2.40 \times 10^{-17} / 9.11 \times 10^{-31}}$
$v = \sqrt{5.27 \times 10^{13}} = 7.26 \times 10^6$ m/s.

Momentum:

$p = m_e v = 9.11 \times 10^{-31} \times 7.26 \times 10^6 = 6.61 \times 10^{-24}$ kg m/s.

De Broglie wavelength:

$\lambda = h / p = 6.63 \times 10^{-34} / 6.61 \times 10^{-24} = 1.00 \times 10^{-10}$ m = 0.10 nm.

Comparison: this is half the typical interatomic spacing, so a crystal acts as an effective diffraction grating for these electrons. This is the basis of electron diffraction (Davisson-Germer, and modern electron microscopes).

Markers reward kinetic energy from $eV$, momentum, de Broglie formula, and the comparison statement (smaller than interatomic spacing, hence diffraction observed).

De Broglie proposed that any particle of momentum $p$ has a wavelength $\lambda = h / p$. Applied to an electron in a circular orbit, the wave goes around the orbit. For a stable, self-reinforcing wave (i.e. a standing wave around the loop), the circumference must contain an integer number of wavelengths:

$2 \pi r = n \lambda$.

Substituting $\lambda = h / p = h / (m_e v)$:

$2 \pi r = n h / (m_e v)$
$m_e v r = n h / (2 \pi) = n \hbar$.

This is exactly Bohr's quantisation condition. The electron in an allowed orbit is a standing wave; non-integer wavelengths would interfere destructively with themselves and cancel.

Markers reward the de Broglie formula, the standing-wave condition around the orbit, and the algebraic identification with Bohr's $m v r = n \hbar$.

Equation. $\lambda = \dfrac{h}{p}$, where $p$ is momentum and $h$ is Planck's constant.

Calculation. $p = m_p v = 1.673 \times 10^{-27} \times 2.0 \times 10^5 = 3.35 \times 10^{-22}\ \text{kg m/s}$.

$\lambda = \dfrac{h}{p} = \dfrac{6.626 \times 10^{-34}}{3.35 \times 10^{-22}} = 2.0 \times 10^{-12}\ \text{m}$.

Marks: one for correctly stating $\lambda = h/p$ and substituting momentum, one for the final answer to two significant figures with the unit metre.

Kinetic energy from the accelerating voltage

$E_K = eV = 1.602 \times 10^{-19} \times 60 = 9.61 \times 10^{-18}\ \text{J}$.

Speed

$v = \sqrt{\dfrac{2E_K}{m_e}} = \sqrt{\dfrac{2 \times 9.61 \times 10^{-18}}{9.109 \times 10^{-31}}} = \sqrt{2.11 \times 10^{13}} = 4.59 \times 10^6\ \text{m/s}$.

Momentum and wavelength

$p = m_e v = 9.109 \times 10^{-31} \times 4.59 \times 10^6 = 4.18 \times 10^{-24}\ \text{kg m/s}$.

$\lambda = \dfrac{h}{p} = \dfrac{6.626 \times 10^{-34}}{4.18 \times 10^{-24}} = 1.6 \times 10^{-10}\ \text{m}$.

Marks: one for $E_K = eV$ correctly evaluated, one for the speed from $v = \sqrt{2E_K/m_e}$, one for $\lambda = h/p = 1.6 \times 10^{-10}$ m with the unit.

Its de Broglie wavelength is $\lambda = \dfrac{h}{mv} = \dfrac{6.626 \times 10^{-34}}{0.060 \times 25} = 4.4 \times 10^{-34}\ \text{m}$.

This is about $20$ orders of magnitude smaller than any aperture, slit spacing or apparatus that exists, so no diffraction or interference effect can ever be produced or detected. The tennis ball genuinely has a matter wavelength; it is just far too small to observe.

Marks: one for calculating $\lambda \approx 4.4 \times 10^{-34}$ m, one for the correct reasoning (wavelength is real but immeasurably smaller than any available structure, not "macroscopic objects have no wave nature").

(a) The graph is a straight line through the origin, so $\lambda$ is directly proportional to $1/\sqrt{V}$ ($\lambda \propto V^{-1/2}$), as predicted by $\lambda = h/\sqrt{2m_e eV}$ for a fixed charge and mass.

(b) Gradient $= \dfrac{\Delta \lambda}{\Delta (1/\sqrt{V})} = \dfrac{(0.245 - 0.123)\ \text{nm}}{(0.200 - 0.100)\ \text{V}^{-1/2}} = \dfrac{0.122\ \text{nm}}{0.100\ \text{V}^{-1/2}} = 1.22\ \text{nm V}^{1/2}$.

(c) The theoretical gradient is $\dfrac{h}{\sqrt{2 m_e e}} = \dfrac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19}}} = \dfrac{6.626 \times 10^{-34}}{5.404 \times 10^{-25}} = 1.226 \times 10^{-9}\ \text{m V}^{1/2} = 1.23\ \text{nm V}^{1/2}$, matching the measured gradient in (b) to three significant figures.

Marks: one for identifying $\lambda \propto 1/\sqrt{V}$ (line through origin), one for a correctly calculated gradient with units of $\text{nm V}^{1/2}$, one for computing the theoretical gradient $h/\sqrt{2m_e e}$, one for the explicit numerical comparison confirming agreement.

(a) Wavelength from the standing-wave condition. $2\pi r = n\lambda$ with $n = 2$, so $\lambda = \dfrac{2\pi r}{n} = \dfrac{2\pi \times 2.12 \times 10^{-10}}{2} = 6.66 \times 10^{-10}\ \text{m}$.

(b) Speed. From $\lambda = h/(m_e v)$, $v = \dfrac{h}{m_e \lambda} = \dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 6.66 \times 10^{-10}} = 1.09 \times 10^{6}\ \text{m/s}$.

Marks: one for applying $2\pi r = n\lambda$ correctly with $n = 2$, one for $\lambda = 6.66 \times 10^{-10}$ m, one for $v = 1.09 \times 10^{6}$ m/s from $\lambda = h/(m_e v)$.

Momentum. $p = \dfrac{h}{\lambda} = \dfrac{6.626 \times 10^{-34}}{0.18 \times 10^{-9}} = 3.68 \times 10^{-24}\ \text{kg m/s}$.

Speed. $v = \dfrac{p}{m_n} = \dfrac{3.68 \times 10^{-24}}{1.675 \times 10^{-27}} = 2.2 \times 10^{3}\ \text{m/s}$.

Marks: one for $p = h/\lambda$ correctly evaluated, one for $v = p/m_n$, one for the final speed $v = 2.2 \times 10^3$ m/s with the unit. (This matches the accepted "thermal neutron" speed of a few km/s, which is why neutron diffraction resolves crystal structure.)

Band-6 plan. (1) State de Broglie's hypothesis and what a confirming experiment would need to show. (2) Describe the Davisson-Germer set-up and the key observation (a sharp diffraction peak, not a smooth scattering curve). (3) Show quantitatively that the observed angle matched the de Broglie wavelength via the Bragg-like condition. (4) Evaluate significance: it made matter waves an experimental fact, not just a postulate, and underpinned the wave mechanics that followed. End with a judgement.

Model answer. De Broglie's 1924 hypothesis proposed that every particle of momentum $p$ has an associated wavelength $\lambda = h/p$, with no experimental support at the time beyond its ability to reproduce Bohr's orbit quantisation. A genuine test required showing that particles produce an interference or diffraction pattern with exactly the predicted wavelength.

Davisson and Germer (1927) fired low-energy electrons at a nickel target that had accidentally crystallised into a single crystal during a vacuum-leak recovery. Scattering the $54\ \text{V}$ electron beam from this crystalline surface produced a sharp intensity peak at $50^{\circ}$, rather than the smooth, featureless scattering expected if electrons behaved as simple classical particles bouncing off atoms. A sharp angular peak is the signature of diffraction: constructive interference from a regularly spaced lattice, described by $d\sin\theta = \lambda$ (a Bragg-type condition).

The wavelength implied by de Broglie's formula for $54\ \text{V}$ electrons is $\lambda = h/\sqrt{2m_eeV} = 6.626 \times 10^{-34}/\sqrt{2 \times 9.109 \times 10^{-31} \times 1.602 \times 10^{-19} \times 54} = 1.67 \times 10^{-10}\ \text{m}$. Using the known nickel spacing $d = 0.215\ \text{nm}$, this predicts a diffraction peak at $\theta = \sin^{-1}(\lambda/d) \approx 51^{\circ}$, matching the observed $50^{\circ}$ to within experimental uncertainty. George Thomson's independent 1927 experiment, passing electrons through thin metal foils to obtain diffraction rings, gave the same conclusion by a different method.

This close numerical agreement, obtained by two independent groups using two different techniques, elevated de Broglie's hypothesis from an elegant postulate to an experimentally verified law. It was decisive evidence that matter possesses genuine wave properties, not merely a mathematical trick for reproducing Bohr's orbits, and it directly motivated Schrodinger's wave mechanics (1926) and the probabilistic interpretation of the electron's position. Without this confirmation, the wave-particle duality at the heart of quantum mechanics would have remained an unverified conjecture.

Marker's note: the top band gives the de Broglie prediction AND the measured angle with a numerical comparison (not just "the results matched"), correctly identifies the sharp peak as the diffraction signature (versus smooth classical scattering), and reaches an explicit judgement of significance (independent confirmation by Thomson; motivated Schrodinger's wave mechanics). A response that only narrates the experiment without the angle/wavelength calculation caps in the middle band.

Band-6 plan. (1) State the stability problem left by Rutherford's model. (2) State de Broglie's hypothesis and the standing-wave condition around an orbit. (3) Derive $m_e v r = n\hbar$ algebraically from $2\pi r = n\lambda$. (4) Evaluate: it explains WHY only certain orbits are stable (a physical mechanism, not an arbitrary rule) but does not itself derive quantised energy levels or explain multi-electron atoms - the full resolution needed Schrodinger. Give a final judgement of "extent".

Model answer. Rutherford's nuclear model placed electrons in orbit around a small dense nucleus, but classical electromagnetism predicts that an accelerating (orbiting) charge continuously radiates energy and should spiral into the nucleus in a fraction of a second. Bohr's 1913 model avoided this by postulating, without physical justification, that only orbits with angular momentum $m_e v r = n\hbar$ ($n = 1, 2, 3, \ldots$) are allowed and that electrons in these orbits do not radiate.

De Broglie's 1924 hypothesis gave this postulate a physical mechanism. If the electron has a wavelength $\lambda = h/p = h/(m_ev)$, then picturing it as a wave travelling around the orbit, the wave can only exist as a stable, self-reinforcing pattern if it closes on itself after one lap - a standing wave, requiring the circumference to be a whole number of wavelengths: $2\pi r = n\lambda$. Substituting $\lambda = h/(m_ev)$ gives $2\pi r = nh/(m_ev)$, which rearranges to $m_evr = nh/(2\pi) = n\hbar$ - exactly Bohr's quantisation condition, now derived rather than assumed. Orbits with a non-integer number of wavelengths would interfere destructively with themselves on successive laps and cannot persist, which is why only the quantised radii are stable.

This is a genuine and significant resolution: it replaces an ad hoc rule with a physical wave-mechanical reason for orbit stability, and the qualitative agreement (electrons only occupy discrete orbits) is confirmed independently by the Davisson-Germer diffraction result. However, the resolution is only partial. De Broglie's standing-wave picture is a simplified, one-dimensional argument that works for circular orbits in hydrogen but does not itself predict the correct energy-level formula for multi-electron atoms, does not account for orbital angular momentum in three dimensions, and still treats the electron as tracing a definite path (a "wave" wrapped around a classical orbit), which the later Schrodinger equation and Heisenberg's uncertainty principle showed is not strictly correct - the electron has no well-defined orbit at all, only a probability distribution.

Weighing this, de Broglie's hypothesis converts Bohr's stability postulate from an assumption into a consequence of a deeper physical principle (wave-particle duality), and this was an essential conceptual bridge to full quantum mechanics; but it does not, by itself, completely resolve atomic structure, since the standing-wave-on-a-circle picture was later superseded by the delocalised orbitals of the Schrodinger equation.

Marker's note: the top band completes the full algebraic derivation ($2\pi r = n\lambda \Rightarrow m_evr = n\hbar$), explicitly states the classical radiation problem being solved, and gives a genuine evaluative judgement of "extent" (real conceptual progress, but not a complete or final theory). A response that only states $\lambda = h/p$ and asserts stability without the derivation caps in the middle band.