Every moving particle has an associated wavelength λ = h/p = h/(mv). For an electron accelerated through potential V: λ = h/√(2meV) = 12.27/√V Å.
-- NCERT Class 12 Physics, Ch. 11, p. 285Matter Waves
8 MCQs3 revision cards9-step worked example
Source: NCERT Dual Nature of Matter and RadiationPYQ coverage: NEET 2020, 2025, 2026Official key: NTA-verifiedLast updated: 23 Sep 2026
Matter Waves, explained for NEET
The wavelength of a moving particle depends on its momentum, not its energy or its speed alone. Aspirants who memorise λ = h/mv and stop there get caught the moment a question gives kinetic energy, or accelerating voltage, or compares two particles of different mass at the same energy.
de Broglie's hypothesis (NCERT Class 12 Physics, Chapter 11, page 283) says every moving material particle has an associated wave of wavelength
λ = h/p = h/(mv)
with h = 6.63 × 10⁻³⁴ J·s. That tiny numerator is why matter waves are invisible for everyday objects: a cricket ball at 20 m/s has λ ≈ 10⁻³⁴ m, far below anything measurable. An electron, being 10⁻³⁰ kg, lands near 10⁻¹⁰ m — comparable to atomic spacing, which is why electrons diffract from crystals and cricket balls do not.
Two re-expressions do most of the work in NEET:
- From kinetic energy: p = √(2mK), so λ = h/√(2mK).
- From accelerating voltage: an electron through V volts gains K = eV, so λ = h/√(2meV) = 12.27/√V Å (NCERT Class 12 Physics, Chapter 11, page 285).
Read those carefully. λ ∝ 1/√K, not 1/K. Doubling the kinetic energy divides the wavelength by √2, not by 2. Doubling the accelerating voltage does the same. This square-root dependence is the single most-tested feature of the relation.
Mass matters too. At equal speed, the heavier particle has the shorter wavelength (λ ∝ 1/m). At equal kinetic energy, the heavier particle still has the shorter wavelength but now λ ∝ 1/√m. At equal momentum, both particles have exactly the same wavelength regardless of mass — that case trips people who reach for mass reflexively.
Watch-out: the formula as written is non-relativistic. If a problem pushes v toward c, h/(mv) with rest mass is no longer valid.
Can you answer these Matter Waves MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The de Broglie wavelength of a moving particle is inversely proportional to its:
Show answer and why every option is right or wrong
Answer: C. C is correct. de Broglie's relation is λ = h/p, so λ varies inversely with momentum p (NCERT Class 12 Physics, Chapter 11, page 283).
Why A is wrong: A is wrong because λ ∝ 1/√K, not 1/K — the kinetic-energy dependence is a square root, since p = √(2mK).
Why B is wrong: B is wrong because the relation is stated in terms of momentum; total energy includes rest energy and does not appear in λ = h/p.
Why D is wrong: D is wrong because λ = h/(mv) contains mass explicitly — two particles at the same speed have different wavelengths if their masses differ.
MCQ 2Direct ApplicationPractice
A proton and an alpha particle move with the same momentum. The ratio of the de Broglie wavelength of the proton to that of the alpha particle is:
Show answer and why every option is right or wrong
Answer: A. A is correct. λ = h/p depends only on momentum, and the two momenta are equal, so the wavelengths are equal (NCERT Class 12 Physics, Chapter 11, page 283).
Why B is wrong: B is wrong because it applies a mass ratio that λ = h/p does not contain; mass enters only when momentum is rewritten in terms of speed or energy.
Why C is wrong: C is wrong because it uses λ ∝ 1/m, which holds at equal speed, not at equal momentum.
Why D is wrong: D is wrong because it combines the alpha-to-proton mass ratio 4 with a mass dependence that is absent when momentum is fixed.
MCQ 3Direct ApplicationPractice
An electron is accelerated from rest through a potential difference of 100 V (exact). Its de Broglie wavelength is closest to:
Show answer and why every option is right or wrong
Answer: B. B is correct. For an electron through V volts, λ = 12.27/√V Å = 12.27/10 Å ≈ 1.2 Å (NCERT Class 12 Physics, Chapter 11, page 285).
Why A is wrong: A is wrong because it divides 12.27 by V instead of by √V, treating the voltage dependence as 1/V.
Why C is wrong: C is wrong because it quotes the numerator 12.27 Å itself, which is the wavelength at V = 1 V, not at 100 V.
Why D is wrong: D is wrong because it multiplies by √V rather than dividing — raising the accelerating voltage shortens the wavelength, it does not lengthen it.
MCQ 4Concept TrapPractice
A cricket ball of mass 0.15 kg travelling at 20 m/s has a de Broglie wavelength of order 10⁻³⁴ m, which no experiment can detect. The reason such a wave is unobservable for macroscopic bodies is that:
Show answer and why every option is right or wrong
Answer: B. B is correct. h ≈ 6.63 × 10⁻³⁴ J·s, so for any macroscopic momentum λ = h/p falls far below the size of any aperture or lattice that could diffract it (NCERT Class 12 Physics, Chapter 11, page 283).
Why A is wrong: A is wrong because de Broglie's hypothesis applies to every moving material particle, not to charged particles only.
Why C is wrong: C is wrong because non-detection is a matter of scale, not of the wave being physically disrupted; the same conclusion holds in vacuum.
Why D is wrong: D is wrong because charge plays no part in λ = h/p — neutrons are neutral and their diffraction is routinely observed.
MCQ 5CalculationPractice
An electron and a proton have the same kinetic energy. Taking the proton to be about 1840 times as massive as the electron, the ratio λ_electron : λ_proton is approximately:
Show answer and why every option is right or wrong
Answer: C. C is correct. At equal K, p = √(2mK), so λ = h/√(2mK) and λ ∝ 1/√m. The ratio is √(m_p/m_e) = √1840 ≈ 43, with the lighter electron having the longer wavelength (NCERT Class 12 Physics, Chapter 11, page 283).
Why A is wrong: A is wrong because it uses λ ∝ 1/m (the equal-speed result) and also assigns the shorter wavelength to the lighter particle.
Why B is wrong: B is wrong because the √1840 factor is right but inverted — at equal energy the electron, being lighter, carries less momentum and so the longer wavelength.
Why D is wrong: D is wrong because it uses the full mass ratio instead of its square root; the equal-energy case gives λ ∝ 1/√m.
MCQ 6CalculationPractice
The de Broglie wavelength of an electron accelerated through a potential difference V is λ. To reduce the wavelength to λ/3, the accelerating potential must be changed to:
Show answer and why every option is right or wrong
Answer: D. D is correct. λ = 12.27/√V Å gives λ ∝ 1/√V, so dividing λ by 3 requires √V to increase threefold, i.e. V to increase ninefold (NCERT Class 12 Physics, Chapter 11, page 285).
Why A is wrong: A is wrong because lowering the potential lowers the momentum and therefore lengthens the wavelength.
Why B is wrong: B is wrong because it both lowers the potential and applies the square incorrectly; V/9 would make the wavelength 3λ.
Why C is wrong: C is wrong because it treats λ ∝ 1/V; tripling V only shortens λ by a factor of √3, not 3.
MCQ 7Easy RecallPractice
In the expression λ = h/(mv) for a material particle, the symbol h denotes a quantity whose SI unit is:
Show answer and why every option is right or wrong
Answer: A. A is correct. h is Planck's constant, 6.63 × 10⁻³⁴ J·s, and λ = h/p is dimensionally consistent only with h in J·s (NCERT Class 12 Physics, Chapter 11, page 283).
Why B is wrong: B is wrong because J·s⁻¹ is the watt, a unit of power; dividing that by momentum would not give a length.
Why C is wrong: C is wrong because the joule is the unit of energy — that is the unit of the product hν, not of h itself.
Why D is wrong: D is wrong because kg·m·s⁻¹ is the unit of momentum p, the denominator of the relation, not of h.
MCQ 8Concept TrapPractice
A beam of electrons and a beam of neutrons travel with the same speed. Which statement about their de Broglie wavelengths is correct?
Show answer and why every option is right or wrong
Answer: C. C is correct. At equal speed, λ = h/(mv) gives λ ∝ 1/m, and the electron is far lighter than the neutron, so its wavelength is much longer (NCERT Class 12 Physics, Chapter 11, page 283).
Why A is wrong: A is wrong because charge does not appear in λ = h/(mv); and the neutron, being heavier, has the shorter wavelength at equal speed.
Why B is wrong: B is wrong because equal speed does not mean equal momentum — mass differs, so p = mv and hence λ differ.
Why D is wrong: D is wrong because the relation is a property of each particle; a beam is simply many such particles, each with the same λ.
Free NEET study resources
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Matter Waves: quick recall before you leave
How do you solve a Matter Waves question? A worked example
- 1
Given
An electron, starting from rest, is accelerated through a potential difference V = 4.00 × 10² V.
Electron mass m = 9.11 × 10⁻³¹ kg; electron charge e = 1.60 × 10⁻¹⁹ C; Planck's constant h = 6.63 × 10⁻³⁴ J·s. - 2
Required
The de Broglie wavelength λ of the electron, in ångström.
- 3
Concept
A particle accelerated from rest through a potential difference V acquires kinetic energy K = eV. Its momentum follows from K = p²/2m, and the de Broglie relation then converts that momentum to a wavelength (NCERT Class 12 Physics, Chapter 11, pages 283 and 285). Because the electron here is well below relativistic speeds, the non-relativistic form is valid.
- 4
Formula
λ = h/p, with p = √(2mK) and K = eV, giving
λ = h/√(2meV) - 5
Substitution
λ = (6.63 × 10⁻³⁴) / √(2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹ × 4.00 × 10²)
- 6
Calculation
Denominator, inside the root: 2 × 9.11 × 10⁻³¹ × 1.60 × 10⁻¹⁹ = 2.92 × 10⁻⁴⁹.
Times 4.00 × 10²: 1.17 × 10⁻⁴⁶.
Square root: √(1.17 × 10⁻⁴⁶) = 1.08 × 10⁻²³ kg·m/s.
λ = (6.63 × 10⁻³⁴)/(1.08 × 10⁻²³) = 6.14 × 10⁻¹¹ m.
The factor 2 in p = √(2mK) is an exact counting number arising from the algebra, not a measurement, so it does not enter the significant-figure count; the three given data (h, m, e) and V all carry three significant figures, so the answer is quoted to three. - 7
Final answer
λ = 6.14 × 10⁻¹¹ m = 0.614 Å.
Cross-check with the shortcut: λ = 12.27/√V Å = 12.27/√(4.00 × 10²) = 12.27/20.0 = 0.614 Å. The two agree, as they must — the shortcut is this same calculation with the constants pre-multiplied. - 8
Common trap
Reaching for λ = h/(mv) and then trying to find v first. That detour works, but it invites a second error: students square-root the energy once to get v, then forget that λ already depends on √V and write λ ∝ 1/V. Equally common is dividing 12.27 by V instead of √V — here that gives 0.031 Å, a factor of 20 out. Go through momentum: p = √(2meV), then λ = h/p. One square root, taken once.
- 9
Similar NEET-style question
An alpha particle (charge 2e, mass 4u) and a proton (charge e, mass 1u) are each accelerated from rest through the same potential difference. Find the ratio λ_proton : λ_alpha.
*(Route: K = qV differs between them, and so does m; use λ = h/√(2mqV) and take the ratio √(m_α q_α / m_p q_p) = √8 = 2√2, so λ_p : λ_α = 2√2 : 1.)*
What to remember before solving Matter Waves questions
Which Matter Waves formulas do you need for NEET?
1 formula — click to collapse
de Broglie wavelength
Wavelength associated with any moving particle of momentum p. For electron through V volts: lambda ~ 12.27/sqrt(V) angstroms.
| Symbol | Quantity | SI Unit |
|---|---|---|
| lambda | de Broglie wavelength | m |
| h | Planck constant | J*s |
| p | momentum | kg*m/s |
| m | mass | kg |
| v | speed | m/s |
Valid when
- Non-relativistic (v << c)
More in Dual Nature of Matter and Radiation: 2 exam traps and mistakes · 2 formulas · 1 question pattern from its other lessons.
Matter Waves questions from past NEET papers
4 questions from NEET 2020, 2025, 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
12/450
21/250
3225
4275
NTA Answer: Option 3(final)
NEET 2025
12.67 nm
20.067 nm
30.67 nm
41.67 nm
NTA Answer: Option 3(final)
NEET 2025
1(1/c)√(E/2m)
2√(E/2m)
3c√(2mE)
4c√(2m/E)
NTA Answer: Option 4(final)
NEET 2020
1104 V
210 V
3102 V
4103 V
NTA Answer: Option 1(final)
All 11 past-paper questions from Dual Nature of Matter and Radiation →
How does NEET ask about Matter Waves?
1 recurring pattern from past papers — click to collapse
de Broglie wavelength λ = h/p. For electron through V volts: λ = 12.27/sqrt(V) Å.
Direct ApplicationEasy
Common distractors
misses relativistic correction
Applies non-relativistic when v ~ c
Sources
NCERT refs: Class 12 Physics Chapter 11, p.283 | Class 12 Physics Chapter 11, p.285
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.
Test yourself on this topic with real past-paper questions:
Practice this topic →