TOPIC 11 OF 27

De Broglie Davisson Germer

🎓 Class 12 Physics CBSE Theory Ch 11 – Dual Nature of Radiation and Matter ⏱ ~14 min
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De Broglie Davisson Germer

11.14 Wave Nature of Matter — A Symmetry Argument

By 1923 the photoelectric effect and Compton scattering had convinced physicists that light, "obviously" a wave, also behaves as a particle. Louis de Broglie, then a doctoral student in Paris, asked the symmetric question: if light shows particle behaviour, do "particles" of matter — electrons, protons, atoms — show wave behaviour too? In his 1924 PhD thesis he proposed that they do.

de Broglie hypothesis (1924): Every moving particle of momentum \(p\) is accompanied by a wave of de Broglie wavelength: \[\boxed{\;\lambda = \frac{h}{p} = \frac{h}{mv}\;}\] The two equations \(E = h\nu\) and \(\lambda = h/p\) together describe both light and matter — the same Planck constant \(h\) ties them.

Why we don't see matter waves in everyday life

Plug everyday numbers into \(\lambda = h/(mv)\):

  • A 60 g cricket ball moving at 20 m/s: \(\lambda = 6.6\times10^{-34}/(0.06\times20) = 5.5\times10^{-34}\) m. Smaller than any nucleus by twenty orders of magnitude — undetectable.
  • An electron (mass \(m_e = 9.1\times10^{-31}\) kg) at \(10^6\) m/s: \(\lambda = 6.6\times10^{-34}/(9.1\times10^{-31}\times10^6) = 7.3\times10^{-10}\) m, comparable to atomic spacing — detectable!

The wave nature of matter dominates only when \(\lambda\) is comparable to the apparatus or the obstacle. For laboratory-scale objects \(\lambda\) is unimaginably small; for electrons it can be made larger than the spacing between atoms in a crystal.

de Broglie wavelength of an accelerated electron

An electron accelerated from rest through potential difference \(V\) gains kinetic energy \(eV\), hence momentum \(p = \sqrt{2m_e eV}\). Its de Broglie wavelength is therefore:

\[\lambda = \frac{h}{\sqrt{2m_e eV}} = \frac{1.227}{\sqrt{V\,(\text{in volts})}}\;\text{nm}\]

So a 100 V electron has \(\lambda \approx 0.123\) nm — comparable to X-ray wavelengths and to the spacing in a crystal.

Accelerating voltage V (V)Electron λ (nm)Compare with
11.227Soft X-ray
100.388Hard X-ray
540.167Used in Davisson-Germer
1000.123Atomic spacing in crystal
10000.039Inside a nucleus
100000.012Electron microscope
A particle (mass m, velocity v) and its de Broglie wave m v → λ = h/(mv) λ
Fig 11.9: Every moving particle has a de Broglie wave of wavelength λ = h/p. The lighter and slower the particle, the longer its λ.

11.15 Davisson–Germer Experiment (1927)

Three years after de Broglie's prediction, Clinton Davisson and Lester Germer, working at Bell Labs in New York, accidentally found electron diffraction while studying electron-surface scattering. Independently, G. P. Thomson in Aberdeen demonstrated electron diffraction through thin metal foils. Both teams confirmed de Broglie's wavelength formula to high precision.

Apparatus

Inside an evacuated chamber:

  • A heated tungsten filament F emits electrons by thermionic emission.
  • A cylindrical anode with a small hole accelerates them through a known potential difference \(V\) (variable, typically 30 V to 100 V).
  • The fine, monoenergetic beam strikes a single crystal of nickel (target T) at normal incidence.
  • A movable electron detector D (a Faraday cylinder) measures the intensity of scattered electrons as a function of the angle ϕ between the incident and scattered directions.
Electron Gun filament F + anode accelerated by V e⁻ beam Ni crystal (target T) normal diffracted at φ φ Detector D (Faraday) D moves through angle φ evacuated chamber
Fig 11.10: Davisson-Germer apparatus — accelerated electrons striking a nickel crystal, scattered intensity measured at angle φ by a movable detector.

The Result — A Diffraction Peak at 50°

Davisson and Germer plotted the intensity of scattered electrons versus the angle φ (between incident and detected beams) for a series of accelerating voltages. At V = 54 V, a sharp peak appeared at φ = 50°. This peak could be explained only as an interference (diffraction) maximum from the regular spacing of nickel atoms — exactly as X-rays would behave.

Treating the surface as a Bragg-like diffraction grating with spacing \(d = 0.215\) nm:

\[\lambda_{\text{exp}} = d\sin\phi = (0.215\,\text{nm})(\sin50°) = 0.165\,\text{nm}\]

Comparing with de Broglie's prediction at 54 V: \(\lambda_{\text{theo}} = 1.227/\sqrt{54} = 0.167\) nm. The two values agreed to better than 1.5%. Matter waves were real.

Ni crystal (incident from above) φ = 0 φ = 50° peak (54 V) Polar plot of scattered intensity I(φ) at V = 54 V
Fig 11.11: Polar plot of scattered electron intensity vs angle φ for V = 54 V. The strong peak at 50° is a diffraction maximum and gives λ ≈ 0.167 nm — confirming de Broglie.

11.16 Wave–Particle Duality — A Two-Sided Coin

The story closes — and modern quantum mechanics opens — with a striking symmetry:

Wave aspectParticle aspect
LightInterference, diffraction, polarisationPhotoelectric effect, Compton scattering
Matter (e.g. electron)Davisson-Germer diffraction; G. P. Thomson; later neutron, atom interferometryMass, charge, kinetic energy, definite trajectory in collisions
Modern interpretation: Neither "wave" nor "particle" fully captures a quantum object. A complete description is given by a wavefunction ψ whose squared magnitude |ψ|² gives the probability density of finding the particle. This is the heart of quantum mechanics, developed by Schrödinger, Heisenberg, Dirac, Born, Pauli and others between 1925 and 1932.
Activity 11.4 — Cricket Ball vs Electron

Compute the de Broglie wavelength for the following objects (all moving at 1 m/s, except where stated). Decide whether a wave aspect could be detected.

ObjectMassSpeed
Cricket ball0.16 kg30 m/s
Bullet0.005 kg500 m/s
Tennis ball0.058 kg1 m/s
Dust grain10⁻⁹ kg1 mm/s
Hydrogen atom1.67×10⁻²⁷ kg1000 m/s
Electron9.11×10⁻³¹ kg1×10⁶ m/s
Predict: Which of these would diffract through the slits in a normal classroom door?
λ = h/(mv): cricket ball ≈ 1.4×10⁻³⁴ m; bullet ≈ 2.6×10⁻³⁴ m; tennis ball ≈ 1.1×10⁻³² m; dust grain ≈ 6.6×10⁻¹⁹ m; H atom ≈ 4.0×10⁻¹⁰ m; electron ≈ 7.3×10⁻¹⁰ m.
None can diffract through a classroom door (~0.8 m). Only the electron and H-atom show wave behaviour, and only when the obstacle has spacing ≲ 1 nm — that is, an atomic crystal.

Interactive — de Broglie Wavelength Calculator

Choose a particle and adjust either its kinetic energy or its accelerating voltage. Read off the de Broglie wavelength and compare with everyday length scales.

Momentum p
de Broglie λ
Compares with
10⁻²⁰ m10⁻¹⁵ m (nucleus)10⁻¹⁰ m (atom)10⁻⁵ m1 m λ

Worked Examples

Example 1 — Electron in TV cathode-ray tube

An electron is accelerated through 1500 V in a CRT. Find its de Broglie wavelength.

\(\lambda = 1.227/\sqrt{V}\) nm = \(1.227/\sqrt{1500}\) nm = 1.227/38.73 = 0.0317 nm. About one-third the diameter of an atom — well within the X-ray range.
Example 2 — Davisson-Germer angle from de Broglie

Predict the diffraction angle φ for 54 V electrons scattering from a Ni crystal of plane spacing d = 0.215 nm. Compare with the experimental value (50°).

\(\lambda = 1.227/\sqrt{54} = 0.167\) nm.
\(\sin\phi = \lambda/d = 0.167/0.215 = 0.776\). \(\phi = 50.9°\) — within 1° of the observed peak.
Example 3 — Wavelength of a thermal neutron

A neutron (mass 1.675×10⁻²⁷ kg) emerges from a nuclear reactor with kinetic energy 0.025 eV (room-temperature thermal energy). Find its de Broglie wavelength.

K = 0.025 eV = 4.0×10⁻²¹ J. \(p = \sqrt{2mK} = \sqrt{2(1.675\times10^{-27})(4.0\times10^{-21})} = 1.16\times10^{-24}\) kg·m/s.
\(\lambda = h/p = 6.626\times10^{-34}/1.16\times10^{-24} = 5.71\times10^{-10}\) m = 0.57 nm — comparable to crystal lattice spacing. This is why thermal neutrons are widely used in neutron-diffraction structure studies.
Example 4 — Why a flying ball doesn't diffract

A 60 g cricket ball moves at 30 m/s. Calculate λ and explain why its wave aspect is undetectable.

\(\lambda = 6.626\times10^{-34}/(0.060\times30) = 3.68\times10^{-34}\) m. To diffract, the ball would need to encounter a slit of comparable width — far smaller than even a proton (10⁻¹⁵ m). No physical apparatus can resolve a wavelength of 10⁻³⁴ m, so the ball appears purely classical.

Competency-Based Questions

Q1. The de Broglie wavelength of a particle of momentum \(p\) is:

  • (a) p/h
  • (b) hp
  • (c) h/p
  • (d) hc/p
(c) λ = h/p (de Broglie's hypothesis).

Q2. The Davisson-Germer experiment used:

  • (a) X-rays scattering off a crystal
  • (b) Electrons scattering off a Ni crystal
  • (c) Neutrons scattering off graphite
  • (d) Photons of UV on a metal
(b) Electrons of 54 eV diffracting from a single nickel crystal.

Q3. (Short Answer) Two particles, an electron and a proton, have the same kinetic energy. Which has the larger de Broglie wavelength, and by what factor?

\(\lambda = h/\sqrt{2mK}\). For the same K, λ ∝ 1/√m. m_p/m_e ≈ 1836, so λ_e/λ_p = √1836 ≈ 42.8. The electron's wavelength is about 43 times larger.

Q4. (Fill in the blank) The wavelength of an electron accelerated through V volts is approximately ______ / √V nm.

1.227 (giving λ in nanometres when V is in volts).

Q5. (HOT) A heavy particle and a light particle have the same de Broglie wavelength. Compare their momenta and their kinetic energies.

Same λ ⇒ same momentum p (since p = h/λ). But K = p²/2m, so the lighter particle has the larger kinetic energy in the ratio m_heavy/m_light.

Assertion–Reason Questions

Options: (A) Both true, R correct explanation of A. (B) Both true, R not the correct explanation. (C) A true, R false. (D) A false, R true.

Assertion: The wave nature of macroscopic objects is never observed in everyday life.

Reason: Their de Broglie wavelengths are vastly smaller than any physical aperture or obstacle they encounter.

(A) Both correct and the reason explains the assertion. λ ∼ 10⁻³⁴ m for everyday masses.

Assertion: The Davisson-Germer experiment gives direct experimental support for de Broglie's hypothesis.

Reason: The angle of the diffraction peak agrees with the wavelength predicted by λ = h/√(2meV).

(A) Both correct and the reason explains the assertion. The 50° peak at V = 54 V matches the predicted λ to within 1%.

Assertion: An electron beam can be used as a diffraction probe of crystal structure.

Reason: The de Broglie wavelength of accelerated electrons is comparable to the inter-atomic spacing in crystals.

(A) Both correct and the reason explains the assertion. This is the basis of low-energy electron diffraction (LEED) and electron microscopy.

Frequently Asked Questions - De Broglie Davisson Germer

What is the main concept covered in De Broglie Davisson Germer?
In NCERT Class 12 Physics Chapter 11 (Dual Nature of Radiation and Matter), "De Broglie Davisson Germer" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is De Broglie Davisson Germer useful in real-life applications?
Real-life applications of "De Broglie Davisson Germer" from NCERT Class 12 Physics Chapter 11 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in De Broglie Davisson Germer?
Key formulas in "De Broglie Davisson Germer" (NCERT Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 11?
NCERT Class 12 Physics Chapter 11 (Dual Nature of Radiation and Matter) is structured so each part builds on the previous one. "De Broglie Davisson Germer" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from De Broglie Davisson Germer?
CBSE board questions from "De Broglie Davisson Germer" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "De Broglie Davisson Germer" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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