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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.

Reviewed by: AI editorial process; not yet individually human-reviewed

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

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=6.626×1034/(0.06×30)=3.7×1034\lambda = 6.626 \times 10^{-34} / (0.06 \times 30) = 3.7 \times 10^{-34} m. No diffraction could ever be observed.
  • Dust speck (m=109m = 10^{-9} kg, v=103v = 10^{-3} m/s): λ6.6×1022\lambda \approx 6.6 \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.

De Broglie wavelength of accelerated electrons versus one over the square root of the accelerating voltage A straight line through the origin rising to the right, showing that the de Broglie wavelength of an electron is directly proportional to one over the square root of the accelerating voltage. Four data points at increasing voltage sit on the line. The gradient equals h over the square root of two times the electron mass times the electronic charge. 1 ∕ √V (V⁻⁰⋅⁵⁵⁵) wavelength λ (nm) 0.050.100.150.20 0.060.120.180.24 gradient = h ∕ √(2mₓe) λ ∝ 1∕√V: a line through the origin.

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.

Standing de Broglie wave around a circular Bohr orbit for n equals four A wavy closed loop traces a circular path around a central nucleus, showing exactly four complete de Broglie wavelengths fitting around the circumference. Because the wave closes on itself after a whole number of wavelengths it reinforces constructively lap after lap, which is why this orbit radius is stable. The radius r and one wavelength lambda are labelled with leader lines. +Ze r λ n = 4: four complete wavelengths close the loop 2πr = nλ, so mₓvr = nħ - Bohr's quantisation, derived from the standing wave.

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 whose surface atomic rows are spaced d=0.215 nmd = 0.215 \text{ nm} apart. 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}. Treating the surface as a diffraction grating, the first-order maximum satisfies dsinθ=λd \sin\theta = \lambda, so sinθ=0.167/0.215=0.777\sin\theta = 0.167 / 0.215 = 0.777, θ=51\theta = 51^{\circ}. The 1927 experiment found peak intensity at a scattering angle of 5050^{\circ}, matching this prediction to within experimental uncertainty - direct confirmation of de Broglie's hypothesis and the wave nature of matter.

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.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksState de Broglie's hypothesis as an equation, and use it to calculate the wavelength of a proton (mp=1.673×1027m_p = 1.673 \times 10^{-27} kg) moving at 2.0×1052.0 \times 10^5 m/s.
Show worked solution →

Equation. λ=hp\lambda = \dfrac{h}{p}, where pp is momentum and hh is Planck's constant.

Calculation. p=mpv=1.673×1027×2.0×105=3.35×1022 kg m/sp = m_p v = 1.673 \times 10^{-27} \times 2.0 \times 10^5 = 3.35 \times 10^{-22}\ \text{kg m/s}.

λ=hp=6.626×10343.35×1022=2.0×1012 m\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 λ=h/p\lambda = h/p and substituting momentum, one for the final answer to two significant figures with the unit metre.

foundation3 marksAn electron is accelerated from rest through a potential difference of 6060 V. Calculate its de Broglie wavelength. (h=6.626×1034h = 6.626 \times 10^{-34} J s, me=9.109×1031m_e = 9.109 \times 10^{-31} kg, e=1.602×1019e = 1.602 \times 10^{-19} C.)
Show worked solution →
Kinetic energy from the accelerating voltage
EK=eV=1.602×1019×60=9.61×1018 JE_K = eV = 1.602 \times 10^{-19} \times 60 = 9.61 \times 10^{-18}\ \text{J}.
Speed
v=2EKme=2×9.61×10189.109×1031=2.11×1013=4.59×106 m/sv = \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=mev=9.109×1031×4.59×106=4.18×1024 kg m/sp = m_e v = 9.109 \times 10^{-31} \times 4.59 \times 10^6 = 4.18 \times 10^{-24}\ \text{kg m/s}.

λ=hp=6.626×10344.18×1024=1.6×1010 m\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 EK=eVE_K = eV correctly evaluated, one for the speed from v=2EK/mev = \sqrt{2E_K/m_e}, one for λ=h/p=1.6×1010\lambda = h/p = 1.6 \times 10^{-10} m with the unit.

foundation2 marksExplain why the wave nature of a thrown tennis ball (m=0.060m = 0.060 kg, v=25v = 25 m/s) is never observed, even though de Broglie's hypothesis applies to it.
Show worked solution →

Its de Broglie wavelength is λ=hmv=6.626×10340.060×25=4.4×1034 m\lambda = \dfrac{h}{mv} = \dfrac{6.626 \times 10^{-34}}{0.060 \times 25} = 4.4 \times 10^{-34}\ \text{m}.

This is about 2020 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 λ4.4×1034\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").

core4 marksThe figure shows the de Broglie wavelength λ\lambda of electrons plotted against 1/V1/\sqrt{V}, where VV is the accelerating voltage. **(a)** Describe the relationship shown. **(b)** Using the points (0.100 V1/2, 0.123 nm)(0.100\ \text{V}^{-1/2},\ 0.123\ \text{nm}) and (0.200 V1/2, 0.245 nm)(0.200\ \text{V}^{-1/2},\ 0.245\ \text{nm}), calculate the gradient. **(c)** Show that this gradient is consistent with λ=h/2meeV\lambda = h/\sqrt{2 m_e e V}.
Show worked solution →

(a) The graph is a straight line through the origin, so λ\lambda is directly proportional to 1/V1/\sqrt{V} (λV1/2\lambda \propto V^{-1/2}), as predicted by λ=h/2meeV\lambda = h/\sqrt{2m_e eV} for a fixed charge and mass.

(b) Gradient =ΔλΔ(1/V)=(0.2450.123) nm(0.2000.100) V1/2=0.122 nm0.100 V1/2=1.22 nm V1/2= \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 h2mee=6.626×10342×9.109×1031×1.602×1019=6.626×10345.404×1025=1.226×109 m V1/2=1.23 nm V1/2\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 λ1/V\lambda \propto 1/\sqrt{V} (line through origin), one for a correctly calculated gradient with units of nm V1/2\text{nm V}^{1/2}, one for computing the theoretical gradient h/2meeh/\sqrt{2m_e e}, one for the explicit numerical comparison confirming agreement.

core3 marksAn electron occupies a Bohr orbit of radius r=2.12×1010r = 2.12 \times 10^{-10} m in which exactly two de Broglie wavelengths fit around the circumference. Calculate (a) the electron's de Broglie wavelength and (b) its speed in this orbit.
Show worked solution →

(a) Wavelength from the standing-wave condition. 2πr=nλ2\pi r = n\lambda with n=2n = 2, so λ=2πrn=2π×2.12×10102=6.66×1010 m\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 λ=h/(mev)\lambda = h/(m_e v), v=hmeλ=6.626×10349.109×1031×6.66×1010=1.09×106 m/sv = \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 2πr=nλ2\pi r = n\lambda correctly with n=2n = 2, one for λ=6.66×1010\lambda = 6.66 \times 10^{-10} m, one for v=1.09×106v = 1.09 \times 10^{6} m/s from λ=h/(mev)\lambda = h/(m_e v).

core3 marksA beam of thermal neutrons (mn=1.675×1027m_n = 1.675 \times 10^{-27} kg) used in a diffraction experiment has a de Broglie wavelength of 0.180.18 nm, comparable to atomic spacings in a crystal. Calculate the neutron's speed and momentum.
Show worked solution →

Momentum. p=hλ=6.626×10340.18×109=3.68×1024 kg m/sp = \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=pmn=3.68×10241.675×1027=2.2×103 m/sv = \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/λp = h/\lambda correctly evaluated, one for v=p/mnv = p/m_n, one for the final speed v=2.2×103v = 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.)

exam6 marksAnalyse how the Davisson-Germer experiment provided direct experimental confirmation of de Broglie's hypothesis, and evaluate its significance for the development of quantum mechanics.
Show worked solution →

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 pp has an associated wavelength λ=h/p\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 V54\ \text{V} electron beam from this crystalline surface produced a sharp intensity peak at 5050^{\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 dsinθ=λd\sin\theta = \lambda (a Bragg-type condition).

The wavelength implied by de Broglie's formula for 54 V54\ \text{V} electrons is λ=h/2meeV=6.626×1034/2×9.109×1031×1.602×1019×54=1.67×1010 m\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 nmd = 0.215\ \text{nm}, this predicts a diffraction peak at θ=sin1(λ/d)51\theta = \sin^{-1}(\lambda/d) \approx 51^{\circ}, matching the observed 5050^{\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.

exam7 marksAssess the extent to which de Broglie's matter-wave hypothesis resolves the problem of Bohr orbit stability that Rutherford's nuclear model could not explain. In your answer, derive the link between the standing-wave condition and Bohr's quantisation of angular momentum.
Show worked solution →

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 mevr=nm_e v r = n\hbar algebraically from 2πr=nλ2\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 mevr=nm_e v r = n\hbar (n=1,2,3,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 λ=h/p=h/(mev)\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: 2πr=nλ2\pi r = n\lambda. Substituting λ=h/(mev)\lambda = h/(m_ev) gives 2πr=nh/(mev)2\pi r = nh/(m_ev), which rearranges to mevr=nh/(2π)=nm_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 (2πr=nλmevr=n2\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 λ=h/p\lambda = h/p and asserts stability without the derivation caps in the middle band.

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