PhysicsCBSE Class 12 · Dual Nature of Radiation and Matter

de Broglie Wavelength Class 12: Matter Waves and λ = h/p

Updated 15 October 2026 · Fact-checked

de Broglie proposed that every moving particle has a wave associated with it, of wavelength λ = h/p = h/mv. In terms of kinetic energy, λ = h/√(2mK); for an electron accelerated through V volt, λ ≈ 1.227/√V nm. The wavelength is noticeable only for tiny particles such as electrons, not for everyday objects.

Wavelength is inversely proportional to mass and speed, so only tiny particles like electrons show noticeable matter waves.

Illustration drawn for QuestionBuddy.

Understand de Broglie Wavelength

Light behaves as a wave in interference and as particles (photons) in the photoelectric effect. In 1924 de Broglie asked the reverse question: if waves can act like particles, can particles act like waves? Nature is symmetric here: moving particles of matter do have wave properties.

The de Broglie relation. A particle of momentum p has an associated matter wave of wavelength $λ = \frac{h}{p} = \frac{h}{m v}$ . The same relation holds for photons ( $p = \frac{h}{λ}$ ), which is what links the two pictures. Note the inverse dependence: a faster or heavier particle has a shorter wavelength.

Wavelength from kinetic energy. Since $K = \frac{p ^{2}}{2 m}$ , $p = 2 m K$ and $λ = \frac{h}{2 m K}$ .

Charged particle accelerated through V. A particle of charge q accelerated from rest through a potential difference V gains $K = q V$ , so $λ = \frac{h}{2 m q V}$ . For an electron this becomes $λ \approx \frac{1.227}{V}$ nm (V in volt), a formula to quote (its derivation is not required for 2026-27). An electron accelerated through 100 V has λ ≈ 0.12 nm, similar to X-ray wavelengths and to the spacing of atoms in crystals; this is why electron beams are diffracted by crystals, and why electron microscopes can show far finer detail than light microscopes.

Why we don't see matter waves of everyday objects. For a 0.15 kg ball at 30 m/s, $λ = \frac{6.63 \times 1 0 ^{- 34}}{4.5} \approx 1.5 \times 1 0^{- 34}$ m, absurdly smaller than any opening or obstacle, so no wave effects (like diffraction) can be observed. Wave nature shows up only when λ is comparable to the size of the obstacle or opening, which happens for electrons, neutrons and atoms.

Comparing particles.

Same speed: λ ∝ 1/m.
Same momentum: same λ.
Same kinetic energy: λ ∝ $\frac{1}{m}$ (an electron has a longer wavelength than a proton of the same K).
Same accelerating voltage: λ ∝ $\frac{1}{m q}$ .

Meaning of the matter wave. The matter wave is not a vibration of the particle's material. Its intensity at a place gives the probability of finding the particle there. A particle with a precisely defined momentum is described by a wave spread out in space; this wave-particle duality is the starting point of quantum mechanics. (The Davisson-Germer experiment, which first confirmed electron waves, is not in the 2026-27 syllabus.)

de Broglie Wavelength: formulas and key facts

de Broglie relation: $λ = \frac{h}{p} = \frac{h}{m v}$; Applies to any moving particle.
In terms of kinetic energy: $λ = \frac{h}{2 m K}$; K in joule.
Charged particle accelerated through V: $λ = \frac{h}{2 m q V}$; K = qV.
Electron accelerated through V volt: $λ \approx \frac{1.227}{V} nm$; Quote directly; derivation not required.
Photon (for comparison): $λ = \frac{h c}{E} \approx \frac{1240}{E ( eV )} nm$; Different from a particle of the same energy.

How to solve de Broglie Wavelength questions

Use this method for de Broglie wavelength numericals and comparisons.

1Identify what is given: speed, momentum, kinetic energy, or accelerating voltage.
2Use λ = h/(mv) for speed, λ = h/p for momentum, λ = h/√(2mK) for kinetic energy (K in joule), λ = h/√(2mqV) for voltage.
3For an electron with V given, use λ ≈ 1.227/√V nm as a check.
4For ratio questions, write λ in terms of the quantity held equal and cancel.
5Compare λ with the size of an obstacle or atomic spacing to decide whether wave effects are noticeable.
6Keep h = 6.63 × 10⁻³⁴ J s, m_e = 9.1 × 10⁻³¹ kg and m_p = 1.67 × 10⁻²⁷ kg handy.

de Broglie Wavelength in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Electron microscope

Suppose an electron microscope accelerates electrons through 50 000 V. (Ignore the small relativistic correction.)

(a) Find the de Broglie wavelength of the electrons. (b) Compare it with a typical visible wavelength of 500 nm. (c) Why does this let the microscope show finer detail?

Try it first, then show the solutionSolution

$λ \approx \frac{1.227}{50000}$ nm $\approx \frac{1.227}{223.6}$ nm $\approx 5.5 \times 1 0^{- 3}$ nm.
$\frac{500}{0.0055} \approx 9 \times 1 0^{4}$ : the electron wavelength is about 90 000 times shorter.
The finest detail an instrument can show is limited by the wavelength used; a much shorter wavelength reveals much smaller structures.

Answer: (a) about 5.5 × 10⁻³ nm; (b) about 90 000 times shorter than visible light; (c) shorter waves resolve finer detail.

Memory hook: Faster electrons, shorter waves, sharper pictures.

Matter waves of everyday objects

Imagine a tiny dust particle of mass $1 0^{- 9}$ kg drifting at 1 mm/s.

(a) Find its de Broglie wavelength. (b) Can any diffraction of this particle be observed? Explain.

Try it first, then show the solutionSolution

$λ = \frac{h}{m v} = \frac{6.63 \times 1 0 ^{- 34}}{1 0 ^{- 9} \times 1 0 ^{- 3}} = 6.63 \times 1 0^{- 22}$ m.
This is far smaller than even an atomic nucleus (about 10⁻¹⁵ m); no opening is small enough to diffract it, so it behaves purely as a particle.

Answer: (a) about 6.6 × 10⁻²² m; (b) no, the wavelength is far too small.

Memory hook: Big mass, tiny wave: everyday objects never look wavy.

Tricks to remember de Broglie Wavelength

Short memory aids. Say each one aloud once, then use it on a question the same day.

Lambda equals h over p
'Light Has Particles, Particles Have Light (waves)': λ = h/p for both.
Use it when: Recalling the relation.
1.227 over root V
Electron through V volt: λ ≈ 1.227/√V nm.
Use it when: Electron numericals.
Same K: lighter is longer
λ ∝ 1/√m for equal kinetic energy: the electron beats the proton.
Use it when: Comparisons.
Four times energy, half the wave
λ ∝ 1/√K.
Use it when: Energy-change questions.

de Broglie Wavelength: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Electron wavelength from speed · 1 mark · Very short answerBoard level

Find the de Broglie wavelength of an electron moving at $1 0^{6}$ m/s.

Try it first, then show the solutionSolution

$λ = \frac{h}{m v} = \frac{6.63 \times 1 0 ^{- 34}}{9.1 \times 1 0 ^{- 31} \times 1 0 ^{6}} \approx 7.3 \times 1 0^{- 10}$ m. (1)

Answer: About 7.3 × 10⁻¹⁰ m (0.73 nm).

Example 2: Why a ball shows no wave nature · 3 marks · Short answerBoard level

(a) State de Broglie's hypothesis. (b) Find the de Broglie wavelength of a 0.15 kg ball moving at 30 m/s, and explain why its wave nature is not observed.

Try it first, then show the solutionSolution

(a) Every moving particle has an associated wave whose wavelength is λ = h/p. (1)
(b) $λ = \frac{6.63 \times 1 0 ^{- 34}}{0.15 \times 30} \approx 1.5 \times 1 0^{- 34}$ m. (1)
Wave effects need openings or obstacles comparable to λ; nothing is remotely that small, so the ball behaves as a particle. (1)

Answer: (a) λ = h/p for every moving particle; (b) about 1.5 × 10⁻³⁴ m, far too small to show wave effects.

Example 3: Electrons accelerated through a voltage · 3 marks · Short answerHarder board level

Find the de Broglie wavelength of an electron accelerated from rest through (a) 100 V and (b) 400 V. (c) By what factor does λ change, and why?

Try it first, then show the solutionSolution

(a) $λ = \frac{h}{2 m e V} = \frac{6.63 \times 1 0 ^{- 34}}{2 \times 9.1 \times 1 0 ^{- 31} \times 1.6 \times 1 0 ^{- 19} \times 100} \approx 1.23 \times 1 0^{- 10}$ m (check: 1.227/√100 nm). (1)
(b) $λ \approx \frac{1.227}{20}$ nm ≈ 0.061 nm. (1)
(c) It halves: λ ∝ 1/√V, and V was multiplied by 4. (1)

Answer: (a) about 0.123 nm; (b) about 0.061 nm; (c) halves, since λ ∝ 1/√V.

Example 4: Proton and electron of equal energy · 2 marks · Very short answerHarder board level

A proton and an electron have the same kinetic energy. Find the ratio of their de Broglie wavelengths, and say which is longer.

Try it first, then show the solutionSolution

$λ = \frac{h}{2 m K}$ , so for equal K, $\frac{λ _{p}}{λ _{e}} = \frac{m _{e}}{m _{p}}$ . (1)
$= \frac{9.1 \times 1 0 ^{- 31}}{1.67 \times 1 0 ^{- 27}} \approx 0.023$ : the electron's wavelength is about 43 times longer. (1)

Answer: λ_p : λ_e ≈ 1 : 43; the electron's is longer.

Example 5: Particles and photons compared · 5 marks · Long answerHOTS / competency

(a) A proton and an alpha particle are accelerated from rest through the same potential difference. Find the ratio of their de Broglie wavelengths. (2)
(b) Compare the wavelength of a 10 keV photon with that of a 10 keV electron. (2)
(c) Which of the two in (b) is better for looking at structures about 0.02 nm in size? (1)

Try it first, then show the solutionSolution

(a) $λ = \frac{h}{2 m q V}$ ; with $m_{α} = 4 m_{p}$ and $q_{α} = 2 e$ : $\frac{λ _{p}}{λ _{α}} = \frac{4 m _{p} \times 2 e}{m _{p} \times e}$ . (1)
$= 8 \approx 2.83$ . (1)
(b) Photon: $λ = \frac{1240}{1 0 ^{4}}$ nm ≈ 0.124 nm. (1)
Electron: $λ \approx \frac{1.227}{1 0 ^{4}}$ nm ≈ 0.0123 nm, about ten times shorter. (1)
(c) The electron, whose wavelength (0.012 nm) is smaller than the structures. (1)

Answer: (a) λ_p : λ_α = 2√2 : 1; (b) photon 0.124 nm, electron 0.0123 nm; (c) the electron.

Example 6: Slow neutrons · 4 marks · Case-basedHarder board level

Read the passage and answer.

Neutrons from a reactor are slowed until their kinetic energy is about 0.025 eV. (m_n = 1.675 × 10⁻²⁷ kg)
(a) Find their momentum. (1)
(b) Find their de Broglie wavelength. (1)
(c) Why are such neutrons useful for studying crystals? (1)
(d) If their kinetic energy were made four times larger, what would their wavelength be? (1)

Try it first, then show the solutionSolution

(a) $p = 2 m K = 2 \times 1.675 \times 1 0^{- 27} \times 0.025 \times 1.6 \times 1 0^{- 19} \approx 3.66 \times 1 0^{- 24}$ kg m/s. (1)
(b) $λ = \frac{h}{p} \approx \frac{6.63 \times 1 0 ^{- 34}}{3.66 \times 1 0 ^{- 24}} \approx 1.8 \times 1 0^{- 10}$ m. (1)
(c) Their wavelength is comparable to the spacing of atoms in crystals, so they are diffracted by crystals. (1)
(d) λ ∝ 1/√K, so it halves to about 0.9 × 10⁻¹⁰ m. (1)

Answer: (a) about 3.7 × 10⁻²⁴ kg m/s; (b) about 0.18 nm; (c) λ matches atomic spacings; (d) about 0.09 nm.

Where students lose marks in de Broglie Wavelength

The slips examiners see most often, and what to write instead.

Mistake: Putting K in eV into λ = h/√(2mK).
Forgetting to convert to joule.
Do this: Multiply eV by 1.6 × 10⁻¹⁹ before using SI formulas.
Usually costs: The answer mark.
Mistake: Using λ = hc/E for an electron.
That is the photon formula.
Do this: For particles use λ = h/p = h/√(2mK).
Usually costs: Most of the numerical.
Mistake: Deriving the electron-wavelength formula at length.
Older notes include the derivation.
Do this: For 2026-27 the derivation is not required; quote λ ≈ 1.227/√V nm or h/√(2meV).
Usually costs: Time.
Mistake: Saying heavier particles have longer wavelengths.
Inverting the relation.
Do this: λ ∝ 1/m at the same speed; heavier means shorter.
Usually costs: 1 mark.

Exam tips

de Broglie wavelength questions appeared in 2023 (55/5/1 Q11 and Q20(a)), 2024 (55/1/1 Q12 and Q20), 2025 (55/1/1 Q10) and 2026 (55/1/1 Q14).
For comparison questions, write λ in terms of the quantity held equal before putting in numbers.
Quote λ = 1.227/√V nm for electrons; it is a quick check on longer calculations.
Explain why everyday objects show no wave nature in one line: λ is far smaller than any obstacle.
The Davisson-Germer experiment and Heisenberg's uncertainty principle are not in the 2026-27 syllabus.

Practice questions on de Broglie Wavelength

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The de Broglie wavelength of an electron accelerated through a potential difference V is proportional to
(a) V (b) √V (c) 1/V (d) 1/√V
Hint
λ = h/√(2meV).
Answer
(d) 1/√V.
Q2 · 1 mark · Very short answerBoard level
Write the de Broglie relation and explain each symbol.
Hint
Wavelength and momentum.
Answer
λ = h/p = h/(mv): λ is the wavelength of the matter wave, h is Planck's constant, p is the particle's momentum (m its mass, v its speed).
Q3 · 1 mark · Very short answerHarder board level
Find the de Broglie wavelength of an electron accelerated through 25 V.
Hint
1.227/√V nm.
Answer
λ ≈ 1.227/5 ≈ 0.245 nm.
Q4 · 1 mark · Very short answerHarder board level
Find the momentum of an electron whose de Broglie wavelength is 0.1 nm.
Hint
p = h/λ.
Answer
p = 6.63 × 10⁻³⁴/10⁻¹⁰ = 6.63 × 10⁻²⁴ kg m/s.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The wave nature of a moving cricket ball cannot be observed.
Reason (R): Its de Broglie wavelength is extremely small compared with any obstacle or opening.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
λ = h/mv with large m.
Answer
(a).
Q6 · 1 mark · Very short answerHOTS / competency
The kinetic energy of a particle is increased four times. What happens to its de Broglie wavelength?
Hint
λ ∝ 1/√K.
Answer
It becomes half.
Q7 · 2 marks · Very short answerHOTS / competency
An electron and a proton have the same de Broglie wavelength. Which has more kinetic energy, and by roughly what factor?
Hint
Same λ means same p; K = p²/2m.
Answer
Same λ means the same momentum. K = p²/(2m), so the lighter electron has more kinetic energy by the factor m_p/m_e ≈ 1836.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.de Broglie wavelength in terms of momentum?Show answer
λ = h/p.
2.In terms of kinetic energy?Show answer
λ = h/√(2mK).
3.For an electron accelerated through V volt?Show answer
λ ≈ 1.227/√V nm.
4.Effect of doubling the speed on λ?Show answer
λ halves.
5.Same K: which has the longer λ, electron or proton?Show answer
The electron.
6.Why don't we observe matter waves of a ball?Show answer
Its wavelength is far too small.

de Broglie Wavelength: frequently asked questions

What are matter waves?

Waves associated with moving particles of matter, proposed by de Broglie. Their wavelength is h/p, and their intensity at a point gives the probability of finding the particle there.

Is the derivation of the electron's de Broglie wavelength in the 2026-27 syllabus?

No. The syllabus covers matter waves and the de Broglie relation. Use λ = h/√(2meV) or 1.227/√V nm without deriving it.

Why are electrons used in microscopes to see very small objects?

Electrons accelerated through thousands of volts have de Broglie wavelengths thousands of times shorter than visible light, so they can resolve much finer detail.

Do photons and electrons of the same energy have the same wavelength?

No. For a photon λ = hc/E; for an electron λ = h/√(2mK). At 10 keV, for example, the electron's wavelength is about ten times shorter than the photon's.

Does a stationary particle have a de Broglie wavelength?

For p = 0 the formula gives an infinite wavelength, meaning no meaningful wave; matter waves are associated with moving particles.