de Broglie: λ = h/(mv). Every moving particle has wave nature. Heisenberg: Δx · Δp ≥ h/(4π); position and momentum cannot both be known precisely.
-- NCERT Class 11 Chemistry, Ch. 2, p. 50Dual Nature de Broglie
8 MCQs3 revision cards9-step worked example
Source: NCERT Structure of AtomOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Dual Nature de Broglie, explained for NEET
De Broglie's hypothesis bridges the classical particle world with quantum wave behaviour: every moving particle has an associated wavelength λ = h/(mv). This relationship, confirmed experimentally by Davisson and Germer's electron diffraction, is the conceptual foundation for why electrons cannot be described by definite orbits in the quantum model (NCERT Class 11 Chemistry Chapter 2, page 52).
The formula and its moving parts. λ = h/(mv), where h is Planck's constant (6.626 × 10⁻³⁴ J·s), m is particle mass in kg, and v is velocity in m/s. Equivalently, λ = h/p where p = mv is momentum. For a particle accelerated through potential V, kinetic energy = eV, so p = √(2meV), giving λ = h/√(2meV).
Where aspirants lose marks. The most common confusion is unit inconsistency: mass given in grams or amu is plugged directly without converting to kg, or velocity in cm/s is used without converting to m/s. Since λ = h/(mv), a factor-of-1000 error in mass gives a factor-of-1000 error in wavelength — and that wrong answer is usually sitting among the options.
A second confusion: applying de Broglie to macroscopic objects and concluding "wavelength is zero." It is not zero — it is negligibly small (≈10⁻³⁴ m for a cricket ball) because mass is enormous. NEET may ask you to compute it to test whether you handle powers of ten correctly.
Key takeaway. Heavier or faster → shorter wavelength. Lighter or slower → longer wavelength. The relationship is inverse: double the momentum, halve the wavelength.
Can you answer these Dual Nature de Broglie MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Direct ApplicationPractice
The de Broglie wavelength of a particle moving with velocity v is λ. If its velocity is doubled, the new wavelength is:
Show answer and why every option is right or wrong
Answer: C. λ = h/(mv). If v → 2v, new λ = h/(m·2v) = λ/2. Wavelength is inversely proportional to velocity.
Why A is wrong: A assumes direct proportionality (λ ∝ v) — the relationship is inverse.
Why B is wrong: B assumes wavelength is independent of velocity — contradicts λ = h/(mv).
Why D is wrong: D confuses λ ∝ 1/v² — there is no squared dependence in de Broglie's relation.
MCQ 2Easy RecallPractice
De Broglie's relationship λ = h/(mv) is valid for:
Show answer and why every option is right or wrong
Answer: D. De Broglie's hypothesis applies to all matter particles (electrons, protons, neutrons, atoms) regardless of charge, as long as they possess momentum (NCERT Class 11 Chemistry Chapter 2, page 52).
Why A is wrong: A incorrectly restricts to charged particles — neutrons also exhibit wave behaviour (confirmed by neutron diffraction).
Why B is wrong: B incorrectly excludes charged particles — electrons were the first to demonstrate de Broglie waves (Davisson-Germer experiment).
Why C is wrong: C confuses photons with matter waves — photons have no rest mass and are described by E = hν, not the de Broglie matter-wave relation.
MCQ 3Direct ApplicationPractice
An electron (mass 9.1 × 10⁻³¹ kg) moves with a velocity of 1.0 × 10⁶ m/s. Its de Broglie wavelength is approximately:
Show answer and why every option is right or wrong
Answer: C. λ = h/(mv) = (6.626 × 10⁻³⁴)/(9.1 × 10⁻³¹ × 1.0 × 10⁶) = 6.626 × 10⁻³⁴ / 9.1 × 10⁻²⁵ ≈ 7.28 × 10⁻¹⁰ m (≈ 7.3 Å).
Why A is wrong: A is wrong: 7.27 × 10⁻⁸ m is 100 times the correct value, which would need a velocity of only 1.0 × 10⁴ m/s.
Why B is wrong: B results from a power-of-ten arithmetic error (using 10⁻³⁰ instead of 10⁻³¹ for electron mass, or mishandling the exponent sum).
Why D is wrong: D is wrong: 7.27 × 10⁻¹² m is 100 times too small, which would need a velocity of 1.0 × 10⁸ m/s.
MCQ 4CalculationPractice
If a proton and an α-particle are accelerated through the same potential difference, the ratio of their de Broglie wavelengths (λ_p / λ_α) is:
Show answer and why every option is right or wrong
Answer: A. For a charged particle accelerated through V: λ = h/√(2mqV). So λ_p/λ_α = √(m_α q_α) / √(m_p q_p) = √(4m_p × 2e) / √(m_p × e) = √8 = 2√2. Hence λ_p : λ_α = 2√2 : 1.
Why B is wrong: B assumes wavelength depends only on potential difference — ignores mass and charge dependence in λ = h/√(2mqV).
Why C is wrong: C accounts for the charge ratio (√2) but forgets the mass ratio (α is 4 times as heavy): √(q_α/q_p) = √2 instead of √(m_α q_α / m_p q_p) = √8.
Why D is wrong: D uses mass ratio directly (4:1) without taking the square root — confuses λ ∝ 1/m with λ ∝ 1/√(mq).
MCQ 5Easy RecallPractice
Which experiment provided the first direct evidence for de Broglie's hypothesis of matter waves?
Show answer and why every option is right or wrong
Answer: B. Davisson and Germer (1927) observed electron diffraction from a nickel crystal, confirming that electrons exhibit wave properties with wavelengths matching de Broglie's prediction.
Why A is wrong: A demonstrated particle nature of light (photons), not wave nature of matter.
Why C is wrong: C demonstrated photon momentum transfer to electrons — particle behaviour of electromagnetic radiation, not wave behaviour of matter.
Why D is wrong: D measured the charge of the electron — no connection to wave properties.
MCQ 6Direct ApplicationPractice
A cricket ball of mass 0.15 kg is bowled at 30 m/s. Its de Broglie wavelength is of the order of:
Show answer and why every option is right or wrong
Answer: A. λ = h/(mv) = 6.626 × 10⁻³⁴ / (0.15 × 30) = 6.626 × 10⁻³⁴ / 4.5 ≈ 1.5 × 10⁻³⁴ m. Order of magnitude: 10⁻³⁴ m.
Why B is wrong: B is 100 times too large: it would need mv ≈ 0.045 kg m/s, but mv = 0.15 × 30 = 4.5 kg m/s.
Why C is wrong: C is the atomic scale (angstrom) — appropriate for electrons, not macroscopic objects. Confuses electron wavelength with ball wavelength.
Why D is wrong: D is an intermediate scale that doesn't arise from any plausible substitution — a guess between atomic and macroscopic.
MCQ 7Concept TrapPractice
An electron and a proton have the same kinetic energy. Which has the longer de Broglie wavelength?
Show answer and why every option is right or wrong
Answer: B. λ = h/√(2mKE). For the same KE, the lighter particle (electron, m_e ≪ m_p) has smaller √(2mKE) and hence larger λ. The electron's wavelength is longer.
Why A is wrong: A reverses the relationship — heavier particles at the same KE have MORE momentum and therefore SHORTER wavelength.
Why C is wrong: C would be true only if both mass and KE were equal — since m_e ≠ m_p, wavelengths differ.
Why D is wrong: D is incorrect because λ = h/√(2mKE) depends on mass and KE; since KE is stated equal, only mass determines which is longer.
MCQ 8CalculationPractice
The de Broglie wavelength of an electron accelerated through a potential difference of 100 V is closest to: (Given: m_e = 9.1 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C, h = 6.6 × 10⁻³⁴ J·s)
Show answer and why every option is right or wrong
Answer: D. λ = h/√(2meV) = 6.6 × 10⁻³⁴ / √(2 × 9.1 × 10⁻³¹ × 1.6 × 10⁻¹⁹ × 100). Denominator: √(2 × 9.1 × 1.6 × 10⁻⁴⁸) = √(29.12 × 10⁻⁴⁸) = 5.4 × 10⁻²⁴. λ ≈ 1.22 × 10⁻¹⁰ m.
Why A is wrong: A is wrong: 1.23 × 10⁻⁶ m is ten thousand times the correct value, about the wavelength of infrared light. An electron accelerated through 100 V has λ ≈ 1.23/√100 nm = 0.123 nm.
Why B is wrong: B results from using V = 1 V instead of 100 V: λ ∝ 1/√V, so the wavelength comes out √100 = 10 times too large.
Why C is wrong: C results from squaring V inside the root (V² = 10⁴ instead of 100), which makes the denominator 10 times too large and λ 10 times too small.
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Dual Nature de Broglie: quick recall before you leave
How do you solve a Dual Nature de Broglie question? A worked example
- 1
Given
• Proton: mass m_p = 1.67 × 10⁻²⁷ kg, charge e = 1.6 × 10⁻¹⁹ C• α-particle: mass m_α = 4m_p = 6.68 × 10⁻²⁷ kg, charge q_α = 2e = 3.2 × 10⁻¹⁹ C• Both accelerated through the same potential difference V
- 2
Required
Ratio λ_p : λ_α
- 3
Concept
For a charged particle starting from rest and accelerated through V: KE = qV. Then λ = h/√(2m·KE) = h/√(2mqV).
- 4
Formula
λ = h / √(2mqV)
- 5
Substitution (ratio form)
λ_p / λ_α = √(m_α · q_α) / √(m_p · q_p) = √(4m_p · 2e) / √(m_p · e) = √(8m_p · e) / √(m_p · e)
- 6
Calculation
λ_p / λ_α = √8 = 2√2 ≈ 2.83
Note: The integers 4 (mass ratio) and 2 (charge ratio) are exact counting/defined values and do not limit significant figures. - 7
Final answer
λ_p : λ_α = 2√2 : 1
The proton has a wavelength 2√2 times longer than the α-particle when both are accelerated through the same potential. - 8
Common trap
Forgetting the charge factor. If you use only the mass ratio: √(m_α/m_p) = √4 = 2, you get the wrong ratio of 2:1 instead of 2√2:1. The α-particle carries charge 2e — this enters the formula via KE = qV.
- 9
Similar NEET-style question
A deuteron (mass 2m_p, charge e) and a proton are accelerated through the same potential difference. Find λ_p/λ_d.
Solution sketch: λ_p/λ_d = √(m_d · q_d) / √(m_p · q_p) = √(2m_p · e) / √(m_p · e) = √2.
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What to remember before solving Dual Nature de Broglie questions
Which Dual Nature de Broglie formulas do you need for NEET?
de Broglie wavelength
Wavelength associated with moving particle of momentum mv.
| Symbol | Quantity | SI Unit |
|---|---|---|
| h | Planck 6.626e-34 | J*s |
| m | mass | kg |
| v | velocity | m/s |
Valid when
- Non-relativistic
More in Structure of Atom: 4 exam traps and mistakes · 4 formulas · 3 question patterns from its other lessons.
Dual Nature de Broglie questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
NCERT refs: Class 11 Chemistry Chapter 2, p.52
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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