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Inquiry Question 3: How is it known that classical physics cannot explain the properties of the atom?

Investigate de Broglie's matter waves, and the experimental evidence that confirms their existence including the Davisson-Germer experiment, and how matter waves explain the stability of Bohr orbits

A focused answer to the HSC Physics Module 8 dot point on de Broglie matter waves. The hypothesis lambda = h/p applied to electrons and to macroscopic objects, the Davisson-Germer electron diffraction experiment, and the standing-wave reinterpretation of Bohr's quantised orbits.

Generated by Claude Opus 4.810 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

NESA wants you to state de Broglie's hypothesis λ=h/p\lambda = h/p for matter waves, use it to calculate wavelengths for electrons and (much smaller) for macroscopic objects, describe the Davisson-Germer experiment as the decisive experimental confirmation, and explain how a standing-wave picture motivates the Bohr quantisation rule.

The answer

De Broglie's hypothesis (1924)

By 1924 the photon picture had established a particle aspect of light (carrying energy hfhf and momentum h/λh/\lambda), even though wave properties were equally well established. Louis de Broglie's PhD thesis proposed the symmetric idea: any particle of momentum pp has an associated wavelength:

λ=hp\boxed{\lambda = \frac{h}{p}}

The same formula that gives the wavelength of light from its photon momentum gives the matter wavelength of any massive particle. For ordinary speeds, p=mvp = m v.

Why no one had noticed

For everyday objects, the wavelength is absurdly small.

  • Tennis ball (m=0.06m = 0.06 kg, v=30v = 30 m/s): λ=6.63×1034/(0.06×30)=4×1034\lambda = 6.63 \times 10^{-34} / (0.06 \times 30) = 4 \times 10^{-34} m. No diffraction could ever be observed.
  • Dust speck (m=109m = 10^{-9} kg, v=103v = 10^{-3} m/s): λ7×1022\lambda \approx 7 \times 10^{-22} m. Still vastly smaller than any apparatus.
  • Electron at 100 eV: λ0.12\lambda \approx 0.12 nm, comparable to atomic spacing. Diffraction observable.
  • Thermal neutron (v2200v \approx 2200 m/s): λ0.18\lambda \approx 0.18 nm. Diffraction observable.

The matter-wave nature shows up only for particles whose wavelength is comparable to some structure they can interact with (crystal lattice spacings, apertures, gratings). For everyday objects the wavelength is too small to ever produce observable interference.

Davisson-Germer experiment (1927)

Clinton Davisson and Lester Germer at Bell Labs were studying low-energy electron scattering from a nickel target. An accident (a vacuum leak followed by a heat treatment that crystallised the nickel) left the surface as a single crystal. Subsequent scattering of electrons at 54 V from the now-crystalline surface showed a sharp angular peak at 50 degrees, exactly where Bragg-like diffraction predicted for the nickel lattice spacing and the de Broglie wavelength of 54 eV electrons (0.167 nm).

Their conclusion: electrons diffract off a crystal in exactly the way X-rays do. The pattern is described by the Bragg condition:

dsinθ=mλd \sin \theta = m \lambda

with λ\lambda given by the de Broglie formula. Plugging in their measured angle and the known nickel spacing returned a wavelength consistent with h/ph/p to high accuracy.

George Thomson (J. J. Thomson's son), independently in 1927, fired electrons through a thin metal foil and obtained ring patterns very similar to those produced by X-rays. The two experiments together confirmed de Broglie's hypothesis.

Matter waves and Bohr orbits

De Broglie immediately applied his hypothesis to the hydrogen atom. Picture the electron as a wave travelling around the nuclear Coulomb potential. For a stable, self-consistent orbit the wave must close on itself (a standing wave around the loop). The circumference must therefore be an integer number of wavelengths:

2πr=nλ=nhp2 \pi r = n \lambda = n \frac{h}{p}

Rearranging:

pr=nh2πp r = n \frac{h}{2 \pi}

That is mevr=nm_e v r = n \hbar, exactly Bohr's quantisation of angular momentum.

So Bohr's third postulate is not an arbitrary rule but a consequence of the electron's wave nature: only orbits whose circumference is a whole number of de Broglie wavelengths support standing waves; all others would destructively interfere with themselves and cancel.

Modern applications

The wave nature of matter is the basis of much of modern science:

  • Electron microscope. Resolves features much smaller than light microscopes can, because the de Broglie wavelength of high-voltage electrons is much shorter than visible light.
  • Neutron diffraction. Used to study magnetic structures in solids and to image hydrogen (which X-rays barely see).
  • Atom interferometry. Cold atoms exhibit matter-wave interference and serve as ultra-sensitive accelerometers and gyroscopes.
  • Quantum mechanics generally. The Schrödinger equation is the wave equation for matter waves, and underlies all of atomic, molecular and solid-state physics.

Try it: De Broglie wavelength calculator to compute matter wavelengths from particle mass, speed (or accelerating voltage for electrons).

Worked example: electron microscope

An electron microscope accelerates electrons through 50 kV. Find the de Broglie wavelength and compare with visible light (550 nm). (Relativistic correction is small here but indicates a real correction at higher voltages.)

Kinetic energy: EK=50×103×1.60×1019=8.0×1015E_K = 50 \times 10^3 \times 1.60 \times 10^{-19} = 8.0 \times 10^{-15} J. (Comparable to mec2=8.2×1014m_e c^2 = 8.2 \times 10^{-14} J, so a relativistic treatment gives a small correction of about 5%; the non-relativistic estimate below is acceptable for HSC.)

Speed: v=2EK/me=1.76×1016=1.3×108v = \sqrt{2 E_K / m_e} = \sqrt{1.76 \times 10^{16}} = 1.3 \times 10^8 m/s. (Relativistic formula would give about 1.24×1081.24 \times 10^8 m/s.)

Momentum: p=mev=9.11×1031×1.3×108=1.2×1022p = m_e v = 9.11 \times 10^{-31} \times 1.3 \times 10^8 = 1.2 \times 10^{-22} kg m/s.

Wavelength: λ=h/p=6.63×1034/1.2×1022=5.5×1012\lambda = h / p = 6.63 \times 10^{-34} / 1.2 \times 10^{-22} = 5.5 \times 10^{-12} m = 5.5 pm.

Compared with 550 nm visible light, the electron wavelength is 100000 times shorter. The smallest features resolvable in the microscope scale roughly with λ\lambda, so the electron microscope resolves features 100000 times smaller than an optical microscope.

Examples in context

Example 1. Electron microscope at the Sydney Microscopy and Microanalysis Centre. A 200 keV scanning transmission electron microscope at the University of Sydney accelerates electrons through V=2.0×105 VV = 2.0 \times 10^5 \text{ V}. Non-relativistic momentum: p=2meV=2×9.11×1031×1.6×1019×2.0×105=2.42×1022 kg m/sp = \sqrt{2 m e V} = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times 2.0 \times 10^5} = 2.42 \times 10^{-22} \text{ kg m/s}. De Broglie wavelength: λ=h/p=6.626×1034/2.42×1022=2.74×1012 m=2.7 pm\lambda = h/p = 6.626 \times 10^{-34} / 2.42 \times 10^{-22} = 2.74 \times 10^{-12} \text{ m} = 2.7 \text{ pm}. (Relativistic correction reduces this to 2.5 pm\sim 2.5 \text{ pm}.) Compare to visible light at 500 nm\sim 500 \text{ nm}: electrons have 200,000×200{,}000 \times shorter wavelength, giving 200,000×200{,}000 \times better diffraction-limited resolution - allowing imaging of individual atoms.

Example 2. Davisson-Germer-style demonstration at UNSW. A 54 V54 \text{ V} electron beam strikes a nickel crystal with d=0.215 nmd = 0.215 \text{ nm} atomic spacing. De Broglie wavelength: λ=h/2meV=6.626×1034/2×9.11×1031×1.6×1019×54=1.67×1010 m=0.167 nm\lambda = h/\sqrt{2 m e V} = 6.626 \times 10^{-34} / \sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times 54} = 1.67 \times 10^{-10} \text{ m} = 0.167 \text{ nm}. Bragg's law nλ=2dsinθn \lambda = 2 d \sin\theta predicts the first-order maximum at sinθ=0.167/(2×0.215)=0.388\sin\theta = 0.167 / (2 \times 0.215) = 0.388, θ=22.8\theta = 22.8^{\circ}. The 1927 experiment found peak intensity at exactly this angle - direct confirmation of de Broglie's hypothesis and the wave nature of matter.

Try this

Q1. Write de Broglie's hypothesis as an equation and state which physical quantities link wave and particle properties. [2 marks]

  • Cue. λ=h/p\lambda = h/p; wavelength (wave) inversely proportional to momentum (particle), with Planck's constant the bridge.

Q2. Calculate the de Broglie wavelength of an electron accelerated through 100 V100 \text{ V}. [3 marks]

  • Cue. p=2meV=2×9.11×1031×1.6×1019×100=5.40×1024 kg m/sp = \sqrt{2 m e V} = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times 100} = 5.40 \times 10^{-24} \text{ kg m/s}; λ=h/p=1.23×1010 m=0.123 nm\lambda = h/p = 1.23 \times 10^{-10} \text{ m} = 0.123 \text{ nm}.

Q3. A neutron from the ANSTO OPAL reactor has KE =0.025 eV= 0.025 \text{ eV} (thermal). (a) Calculate the neutron's de Broglie wavelength. (b) Compare this with the spacing between atoms in a solid (0.2 nm\sim 0.2 \text{ nm}). (c) Explain why neutron diffraction is used to study atomic and magnetic structure. [3+2+2 marks]

  • Cue. (a) p=2mnEk=2×1.67×1027×4×1021=3.66×1024p = \sqrt{2 m_n E_k} = \sqrt{2 \times 1.67 \times 10^{-27} \times 4 \times 10^{-21}} = 3.66 \times 10^{-24}; λ=1.81×1010 m\lambda = 1.81 \times 10^{-10} \text{ m}. (b) Comparable, ideal for diffraction. (c) Strong nuclear/magnetic scattering, no electronic interaction.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 HSC5 marksAn electron is accelerated from rest through a potential difference of 150 V. Calculate its de Broglie wavelength and compare it with the typical interatomic spacing in a crystal of 0.2 nm. (h = 6.63 x 10^-34 J s, m_e = 9.11 x 10^-31 kg, e = 1.60 x 10^-19 C.)
Show worked answer →

Kinetic energy gained: EK=eV=1.60×1019×150=2.40×1017E_K = eV = 1.60 \times 10^{-19} \times 150 = 2.40 \times 10^{-17} J.

Speed (non-relativistic at this energy):

v=2EK/me=2×2.40×1017/9.11×1031v = \sqrt{2 E_K / m_e} = \sqrt{2 \times 2.40 \times 10^{-17} / 9.11 \times 10^{-31}}
v=5.27×1013=7.26×106v = \sqrt{5.27 \times 10^{13}} = 7.26 \times 10^6 m/s.

Momentum:

p=mev=9.11×1031×7.26×106=6.61×1024p = 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:

λ=h/p=6.63×1034/6.61×1024=1.00×1010\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 eVeV, momentum, de Broglie formula, and the comparison statement (smaller than interatomic spacing, hence diffraction observed).

2020 HSC4 marksExplain how de Broglie's hypothesis accounts for the quantisation of angular momentum in Bohr's model of the hydrogen atom.
Show worked answer →

De Broglie proposed that any particle of momentum pp has a wavelength λ=h/p\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:

2πr=nλ2 \pi r = n \lambda.

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

2πr=nh/(mev)2 \pi r = n h / (m_e v)
mevr=nh/(2π)=nm_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 mvr=nm v r = n \hbar.

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