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. 24Dual Nature De Broglie
Lesson
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 24).
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.
Practice MCQs
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The de Broglie wavelength of a particle moving with velocity v is λ. If its velocity is doubled, the new wavelength is:
De Broglie's relationship λ = h/(mv) is valid for:
An electron (mass 9.1 × 10⁻³¹ kg) moves with a velocity of 1.0 × 10⁶ m/s. Its de Broglie wavelength is approximately:
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:
Which experiment provided the first direct evidence for de Broglie's hypothesis of matter waves?
A cricket ball of mass 0.15 kg is bowled at 30 m/s. Its de Broglie wavelength is of the order of:
An electron and a proton have the same kinetic energy. Which has the longer de Broglie wavelength?
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)
Quick recall before you leave
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. ---
Before solving, remember these
Formulas
Bohr energy (hydrogen-like)
Energy of nth orbit. Negative (bound). Ground state H: -13.6 eV.
| Symbol | Quantity | SI Unit |
|---|---|---|
| E_n | orbit energy | eV |
| Z | nuclear charge | - |
| n | principal | - |
Valid when
- Hydrogen-like atom
- Non-relativistic
Bohr radius (hydrogen-like)
Radius of nth Bohr orbit for hydrogen-like atom of nuclear charge Z.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | principal quantum number | - |
| Z | nuclear charge | - |
| r_n | orbit radius | Å |
Valid when
- Hydrogen-like (one-electron) atom
- Non-relativistic
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
Heisenberg uncertainty
Position and momentum cannot both be known with arbitrary precision.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Δx | position uncertainty | m |
| Δp | momentum uncertainty | kg*m/s |
Valid when
- Quantum scale; meaningful only when Δx, Δp comparable to atomic dimensions
Rydberg formula (H spectrum)
Spectral wavelengths of hydrogen-like atoms. Lyman (n1=1, UV), Balmer (n1=2, visible), Paschen (n1=3, IR).
| Symbol | Quantity | SI Unit |
|---|---|---|
| lambda | wavelength | m |
| R_H | Rydberg 1.097e7 | 1/m |
| Z | nuclear charge | - |
| n1, n2 | integers, n2>n1 | - |
Valid when
- One-electron atom
Exam Traps & Common Mistakes
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Inorganic Exception
Student writes Cr as [Ar]3d⁴4s² (expected) instead of actual [Ar]3d⁵4s¹. Same for Cu: actual [Ar]3d¹⁰4s¹ (one e⁻ promoted from 4s to 3d).
When it triggers
Question asks for ground-state electronic configuration of Cr (Z=24) or Cu (Z=29).
How to avoid
Half-filled (d⁵) and fully filled (d¹⁰) configurations have extra stability from exchange energy and symmetry. Cr and Cu adopt these configurations by promoting one 4s electron.
Category: Similar Terms
Student forgets Z² scaling when applying Bohr formulas to He⁺ (Z=2) or Li²⁺ (Z=3).
When it triggers
Question involves hydrogen-like ion (He+, Li2+, etc.).
How to avoid
E_n = -13.6 × Z²/n² eV. r_n = (0.529/Z) × n² Å. He+: 4× more bound than H. Li²⁺: 9× more bound. Always include Z².
Root cause: concept gap
Correction
Half-filled (d⁵) and full-filled (d¹⁰) configurations have extra exchange-energy stability. Cr and Cu adopt these by promoting one 4s electron.
Root cause: formula misuse
Correction
Always include Z². E_n = -13.6 × Z²/n². For He+: 4× more energetic than H. For Li²⁺: 9×.
Past Year Questions
9 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
From the following pairs of ions which one is not an iso-electronic pair? Fe2+, Mn2+
How NEET usually asks this
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Write electronic configuration, especially handling Cr/Cu anomalies and ions.
Common distractors
misses cr cu anomaly
Writes 3d⁴4s² instead of 3d⁵4s¹
Bohr energy difference between two levels; compute photon energy/wavelength using Rydberg or E_n formulas.
Common distractors
forgets z squared
Drops Z² for hydrogen-like
Apply Δx·Δp ≥ h/(4π) to find minimum uncertainty given the other.
Common distractors
uses h instead of h over 4pi
Drops 4π factor
Given orbital (e.g. 3p_z) or electron, identify n, l, m_l, m_s. Or check forbidden quantum number combinations.
Common distractors
uses wrong l range
Uses l ≤ n instead of l < n
Sources
Test yourself on this topic with real past-paper questions:
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