PhysicsISC Class 12 · Dual Nature of Radiation and Matter

De Broglie Wavelength: Matter Waves, Formulas and the Davisson-Germer Experiment

Updated 15 October 2026 · Fact-checked

De Broglie proposed that every moving particle has a wave of wavelength $\lambda = \frac{h}{p} = \frac{h}{mv}$. In terms of kinetic energy K, $\lambda = \frac{h}{\sqrt{2mK}}$; for a charge q accelerated through V, $\lambda = \frac{h}{\sqrt{2mqV}}$. For an electron this is $\lambda \approx \frac{1.227}{\sqrt{V}}$ nm. Electron diffraction confirms it.

Understand De Broglie Wavelength and Matter Waves

De Broglie's hypothesis. Light behaves as a wave (interference, diffraction) and as particles (photons, with $p = \frac{h}{λ}$ ). In 1924 de Broglie argued, from the symmetry of nature, that particles such as electrons should also behave as waves, with
$λ = \frac{h}{p} = \frac{h}{m v} .$
These matter waves are not electromagnetic; they describe where the particle is likely to be found.

Forms of the formula. Using $K = \frac{p ^{2}}{2 m}$ :
$λ = \frac{h}{2 m K} .$
A particle of charge q accelerated from rest through potential difference V gains K = qV, so
$λ = \frac{h}{2 m q V} .$
For an electron ( $m = 9.1 \times 1 0^{- 31}$ kg, $q = 1.6 \times 1 0^{- 19}$ C):
$λ = \frac{6.63 \times 1 0 ^{- 34}}{2 \times 9.1 \times 1 0 ^{- 31} \times 1.6 \times 1 0 ^{- 19} \times V} \approx \frac{1.227}{V} nm .$

Proportionalities to use in ratio questions.

Same speed: $λ \propto \frac{1}{m}$ .
Same momentum: same λ, whatever the particle.
Same kinetic energy: $λ \propto \frac{1}{m}$ .
Same accelerating voltage: $λ \propto \frac{1}{m q}$ .
For one particle, $λ \propto \frac{1}{K}$ : making K four times halves λ.

Why we do not notice matter waves in daily life. h is tiny ( $6.63 \times 1 0^{- 34}$ J s). For everyday masses and speeds, λ is far smaller than any atom or aperture, so no diffraction can be seen. For electrons accelerated through tens or hundreds of volts, λ is about 0.1 nm, comparable to the spacing of atoms in a crystal, so crystals act as diffraction gratings.

Davisson-Germer experiment (qualitative). A beam of electrons accelerated through a known voltage was directed at a nickel crystal, and a detector measured the number of electrons scattered at different angles. At 54 V a sharp peak in intensity appeared at a scattering angle of 50°. This peak is a diffraction maximum from the regular rows of atoms. The wavelength found from the crystal spacing, about 0.165 nm, agreed with de Broglie's prediction $\frac{1.227}{54} \approx 0.167$ nm. Conclusion: electrons have a wave nature, and $λ = \frac{h}{p}$ is correct. Electron diffraction (or interference) is the phenomenon that shows particles behaving as waves.

Uses. The electron microscope uses electrons of very short λ (picometres at tens of kilovolts), so it resolves detail thousands of times finer than a light microscope. Neutron diffraction, with slow neutrons of λ ≈ 0.1 nm, is used to study crystal structures.

De Broglie Wavelength and Matter Waves: formulas and key facts

De Broglie relation: $λ = \frac{h}{p} = \frac{h}{m v}$; Any particle.
From kinetic energy: $λ = \frac{h}{2 m K}$; Non-relativistic.
From accelerating voltage: $λ = \frac{h}{2 m q V}$; Starts from rest.
Electron shortcut: $λ \approx \frac{1.227}{V} nm$; V in volts.
Photon momentum: $p = \frac{h}{λ} = \frac{E}{c}$; For comparison.

How to solve De Broglie Wavelength and Matter Waves questions

Choose the form that matches the data: speed, kinetic energy or voltage.

1Speed given: $λ = \frac{h}{m v}$ .
2Kinetic energy given: convert eV to J (× $1.6 \times 1 0^{- 19}$ ), then $λ = \frac{h}{2 m K}$ .
3Voltage given (electron): $λ = \frac{1.227}{V}$ nm; other particles: $\frac{h}{2 m q V}$ .
4Ratio questions: write λ in the relevant form and cancel what is the same.
5Reverse problems: from λ find p, then K or V.

De Broglie Wavelength and Matter Waves 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.

A cricket ball as a wave?

Suppose a fast bowler delivers a 0.16 kg ball at 40 m/s.

Find its de Broglie wavelength and say whether its wave nature could ever be detected.

Try it first, then show the solutionSolution

$λ = \frac{6.63 \times 1 0 ^{- 34}}{0.16 \times 40} \approx 1.04 \times 1 0^{- 34}$ m.
This is about $1 0^{- 19}$ times the size of a nucleus; no gap or obstacle is small enough to diffract it.

Answer: ≈ $1 0^{- 34}$ m: undetectable, so the ball behaves as a particle.

Memory hook: Big mass, tiny wave.

Seeing a virus

Imagine an electron microscope in a hospital laboratory accelerating electrons through 20 kV to image viruses (ignore relativity).

Find the electrons' wavelength and compare it with green light of 550 nm.

Try it first, then show the solutionSolution

$λ = \frac{1.227}{20 000} \approx 8.7 \times 1 0^{- 3}$ nm (8.7 pm).
$\frac{550}{8.7 \times 1 0 ^{- 3}} \approx 6 \times 1 0^{4}$ : the electron wavelength is tens of thousands of times shorter.

Answer: ≈ 8.7 pm; resolution can be far finer than with light.

Memory hook: Shorter wave, sharper picture.

Tricks to remember De Broglie Wavelength and Matter Waves

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

h over p
De Broglie: "Wavelength is Planck over momentum".
Use it when: Basic relation.
Root two m K
$p = 2 m K$ , and with K = qV, $p = 2 m q V$ .
Use it when: Switching forms.
1.227 over root V
Electron only, answer in nm.
Use it when: Fast electron wavelengths.
Four times K, half the λ
$λ \propto \frac{1}{K}$ .
Use it when: Ratio questions (2025).

De Broglie Wavelength and Matter Waves: solved examples

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

Example 1: Doubling momentum · 1 mark · MCQBoard level

If the momentum of a particle is doubled, its de Broglie wavelength:
(a) doubles
(b) halves
(c) becomes four times
(d) is unchanged

Try it first, then show the solutionSolution

$λ = \frac{h}{p}$ , so λ ∝ $\frac{1}{p}$ .

Answer: (b).

Example 2: Electron through 150 V · 2 marks · Short answerBoard level

Find the de Broglie wavelength of an electron accelerated from rest through 150 V.

Try it first, then show the solutionSolution

$λ = \frac{1.227}{V}$ nm for an electron. (1)
$λ = \frac{1.227}{150} \approx 0.100$ nm $= 1.0 \times 1 0^{- 10}$ m. (1)

Answer: ≈ 0.10 nm.

Example 3: Deriving the voltage form · 3 marks · Long answerHarder board level

Show that the de Broglie wavelength of a particle of mass m and charge q accelerated from rest through V is $λ = \frac{h}{2 m q V}$ , and hence that for an electron $λ \approx \frac{1.227}{V}$ nm.

Try it first, then show the solutionSolution

Work done by the field = kinetic energy gained: $q V = \frac{p ^{2}}{2 m}$ , so $p = 2 m q V$ . (1)
$λ = \frac{h}{p} = \frac{h}{2 m q V}$ . (1)
For an electron: $2 \times 9.1 \times 1 0^{- 31} \times 1.6 \times 1 0^{- 19} \approx 5.40 \times 1 0^{- 25}$ , so $λ = \frac{6.63 \times 1 0 ^{- 34}}{5.40 \times 1 0 ^{- 25} V} \approx \frac{1.227 \times 1 0 ^{- 9}}{V}$ m. (1)

Answer: $λ = \frac{h}{2 m q V} \approx \frac{1.227}{V}$ nm.

Example 4: Thermal neutrons · 3 marks · Short answerHarder board level

Slow (thermal) neutrons used in neutron diffraction have kinetic energy about 0.025 eV. Find their de Broglie wavelength ( $m_{n} = 1.675 \times 1 0^{- 27}$ kg) and explain why they suit crystal studies.

Try it first, then show the solutionSolution

$K = 0.025 \times 1.6 \times 1 0^{- 19} = 4.0 \times 1 0^{- 21}$ J. (1)
$p = 2 m K = 2 \times 1.675 \times 1 0^{- 27} \times 4.0 \times 1 0^{- 21} \approx 3.66 \times 1 0^{- 24}$ kg m/s; $λ = \frac{h}{p} \approx 1.81 \times 1 0^{- 10}$ m. (1)
About 0.18 nm, comparable to atomic spacings in crystals, so crystals diffract them strongly. (1)

Answer: ≈ 0.18 nm, matching interatomic spacing.

Example 5: Proton or alpha particle? · 2 marks · Short answerHOTS / competency

A proton and an alpha particle (mass 4 times, charge 2 times that of the proton) are accelerated from rest through the same potential difference. Find the ratio of their de Broglie wavelengths.

Try it first, then show the solutionSolution

$λ \propto \frac{1}{m q}$ for the same V. (1)
$\frac{λ _{p}}{λ _{α}} = \frac{4 m \times 2 e}{m \times e} = 8 = 22 \approx 2.83$ . (1)

Answer: $λ_{p} : λ_{α} = 22 : 1$ .

Example 6: Davisson and Germer · 4 marks · Case-basedHOTS / competency

Suppose you repeat the Davisson-Germer experiment: electrons accelerated through 54 V strike a nickel crystal, and the detector shows a strong peak at a scattering angle of 50°.
(i) Find the de Broglie wavelength predicted for these electrons. (1 mark)
(ii) The wavelength calculated from the crystal spacing is 0.165 nm. Compare. (1 mark)
(iii) What does the sharp peak at one angle show? (1 mark)
(iv) Name the phenomenon that this experiment demonstrates. (1 mark)

Try it first, then show the solutionSolution

(i) $λ = \frac{1.227}{54} \approx 0.167$ nm.
(ii) 0.165 nm and 0.167 nm agree closely, confirming $λ = \frac{h}{p}$ .
(iii) Electrons scattered from regular rows of atoms reinforce at one angle: a diffraction maximum, which only waves produce.
(iv) Electron diffraction, showing the wave nature of electrons.

Answer: (i) ≈ 0.167 nm; (ii) agreement; (iii) a diffraction maximum; (iv) electron diffraction.

Where students lose marks in De Broglie Wavelength and Matter Waves

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

Mistake: Choosing the wrong form
The ISC 2024 examiners (Q1(A)(vi)) reported that distractor options, such as one with an extra factor of 2, caught candidates who did not know the formula well.
Wrong working
$λ = \frac{2 h}{2 m E}$ .
Right working
$λ = \frac{h}{2 m E}$ .
Do this: Start from $λ = \frac{h}{p}$ and $E = \frac{p ^{2}}{2 m}$ .
Usually costs: 1 mark.
Mistake: Ratio slips
In 2025 Q1(A)(iii) candidates did not know how λ changes when K is made four times.
Do this: λ ∝ $\frac{1}{K}$ : four times K gives half λ.
Usually costs: 1 mark.
Mistake: Naming the experiment for the phenomenon
In 2025 Q1(B)(v) candidates wrote "Davisson-Germer experiment" or "de Broglie waves" when asked for a phenomenon.
Do this: Answer "electron diffraction".
Usually costs: 1 mark.
Mistake: Using 1.227/√V for protons
The shortcut holds only for electrons.
Do this: For other particles use $\frac{h}{2 m q V}$ .
Usually costs: 1 mark.

Exam tips

State the starting relation $λ = \frac{h}{p}$ in every numerical.
Convert eV to joules before using $2 m K$ .
In ratio questions, write what is common and cancel it before substituting.
Keep Davisson-Germer answers qualitative: set-up, peak at 54 V and 50°, conclusion.

Practice questions on De Broglie Wavelength and Matter Waves

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
If the kinetic energy of an electron is made nine times, its de Broglie wavelength becomes:
(a) 9λ
(b) 3λ
(c) $\frac{λ}{3}$
(d) $\frac{λ}{9}$
Hint
$λ \propto \frac{1}{K}$ .
Answer
(c).
Q2 · 1 mark · Short answerBoard level
Find the de Broglie wavelength of an electron moving at $2.0 \times 1 0^{6}$ m/s.
Hint
$\frac{h}{m v}$ .
Answer
≈ $3.64 \times 1 0^{- 10}$ m.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The wave nature of a moving cricket ball is not observed.
Reason (R): Its de Broglie wavelength is far smaller than any obstacle or aperture.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Diffraction needs λ ≈ size.
Answer
(a).
Q4 · 2 marks · Short answerHarder board level
Through what potential difference must an electron be accelerated to have a de Broglie wavelength of 0.05 nm?
Hint
$V = (\frac{1.227}{λ})^{2}$ .
Answer
≈ 602 V.
Q5 · 2 marks · Short answerHarder board level
Name the phenomenon in which moving particles behave like waves, and state the conclusion of the Davisson-Germer experiment.
Hint
Not "de Broglie waves".
Answer
Electron diffraction (or interference). Davisson and Germer found a diffraction peak whose wavelength matched $\frac{h}{p}$ , confirming the wave nature of electrons.
Q6 · 3 marks · Short answerHOTS / competency
A photon and an electron each have a wavelength of 0.2 nm. Find the photon's energy and the electron's kinetic energy, in eV.
Hint
Photon: $\frac{1240}{λ}$ ; electron: $(\frac{1.227}{λ})^{2}$ .
Answer
Photon ≈ 6200 eV; electron ≈ 37.6 eV. Same momentum, very different energies.
Q7 · 1 mark · Short answerHOTS / competency
A speck of dust of mass 1 µg moves at 1 mm/s. Find its de Broglie wavelength.
Hint
1 µg = $1 0^{- 9}$ kg.
Answer
≈ $6.6 \times 1 0^{- 22}$ m.
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose an electron and a proton are each accelerated from rest through 200 V ( $\frac{m _{p}}{m _{e}} = 1836$ ).
(i) Find the electron's wavelength. (1 mark)
(ii) Which has the shorter wavelength? (1 mark)
(iii) Find the ratio $\frac{λ _{e}}{λ _{p}}$ . (1 mark)
(iv) Find the proton's wavelength. (1 mark)
Hint
Same charge and V: λ ∝ $\frac{1}{m}$ .
Answer
(i) ≈ 0.0868 nm. (ii) The proton (heavier, same kinetic energy). (iii) $1836 \approx 42.9$ . (iv) ≈ $2.03 \times 1 0^{- 3}$ nm.

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 relation?Show answer
$λ = \frac{h}{p}$ .
2.λ in terms of kinetic energy?Show answer
$\frac{h}{2 m K}$ .
3.Electron at 100 V: λ?Show answer
≈ 0.123 nm.
4.Voltage increased 100 times: λ?Show answer
Becomes one-tenth.
5.Phenomenon showing wave nature of electrons?Show answer
Electron diffraction.

De Broglie Wavelength and Matter Waves: frequently asked questions

Are matter waves electromagnetic?

No. They are associated with any moving particle, charged or not, and are not radiated like light.

Why does an electron microscope see finer detail?

Resolution is limited by wavelength, and electrons accelerated through kilovolts have wavelengths of picometres, far shorter than visible light.

Do particles at rest have a de Broglie wavelength?

At p = 0 the formula gives an infinite wavelength; the idea is useful only for moving particles.

What does the ISC 2027 syllabus need here?

De Broglie's hypothesis, \lambda = \frac{h}{p} = \frac{h}{mv}, electron diffraction (qualitative) and a qualitative description of the Davisson-Germer experiment and its conclusion.

How was this asked recently?

In 1-mark items: ISC 2024 Q1(A)(vi) (λ in terms of kinetic energy), ISC 2025 Q1(A)(iii) (K made four times) and Q1(B)(v) (a phenomenon showing wave nature of particles).