de Broglie Hypothesis: When Particles Ride Waves
The Ant-Man Quantum Realm Connection 🎬
In Ant-Man, Scott Lang shrinks to quantum scales where classical physics breaks down. At those scales, even particles like electrons behave like waves! This isn’t science fiction - it’s the de Broglie hypothesis that revolutionized quantum mechanics.
“If light can be both wave and particle, why not matter?” - Louis de Broglie, 1924
The Historical Context
Einstein’s Photon (1905)
Light waves → Particles (photons)
$$E = hν, \quad p = \frac{h}{λ}$$de Broglie’s Bold Leap (1924)
If waves can be particles… Can particles be waves? 🤔
de Broglie’s Hypothesis
Every moving particle has an associated wave called a matter wave or de Broglie wave.
:::box The de Broglie Wavelength
$$λ = \frac{h}{p} = \frac{h}{mv}$$where:
- λ = de Broglie wavelength
- h = 6.626 × 10⁻³⁴ J·s (Planck’s constant)
- p = momentum of particle
- m = mass, v = velocity :::
Key Insight: Symmetry in Nature
| Particle Property | Wave Property |
|---|---|
| Energy: E = mc² | Frequency: ν = E/h |
| Momentum: p = mv | Wavelength: λ = h/p |
Multiple Forms of de Broglie Equation
For a Free Particle
$$λ = \frac{h}{mv}$$In Terms of Kinetic Energy
$$K = \frac{1}{2}mv^2 = \frac{p^2}{2m}$$ $$p = \sqrt{2mK}$$:::box
$$λ = \frac{h}{\sqrt{2mK}}$$:::
Interactive Demo: Visualize Matter Waves
Watch how particles exhibit wave-like behavior with de Broglie wavelength.
For Charged Particles in Electric Field
When a particle of charge q is accelerated through potential V:
$$K = qV$$:::box
$$λ = \frac{h}{\sqrt{2mqV}}$$:::
For electrons (q = e):
$$λ = \frac{h}{\sqrt{2meV}} = \frac{12.27}{\sqrt{V}} \text{ Å}$$where V is in volts.
Relativistic Form (for high speeds)
$$λ = \frac{h}{\sqrt{2m_0K(1 + K/2m_0c^2)}}$$Wave-Particle Duality: The Complete Picture
Electromagnetic Radiation
- Wave nature: Interference, diffraction, polarization
- Particle nature: Photoelectric effect, Compton effect
Material Particles
- Particle nature: Definite mass, charge, trajectory
- Wave nature: Interference, diffraction (Davisson-Germer)
de Broglie’s Unification
DUALITY
↓
┌──────┴──────┐
↓ ↓
WAVES PARTICLES
↓ ↓
λ = c/ν p = mv
↓ ↓
└──── h ─────┘
↓
λ = h/p
Phase Velocity vs Group Velocity
Phase Velocity (vₚ)
Velocity of a single wave component:
$$v_p = νλ = \frac{ω}{k}$$For matter waves:
$$v_p = \frac{E}{p} = \frac{mc^2}{mv} = \frac{c^2}{v}$$Note: vₚ > c (but no information travels at vₚ!)
Group Velocity (vₘ)
Velocity of wave packet (actual particle velocity):
$$v_g = \frac{dω}{dk}$$For matter waves: vₘ = v (particle velocity)
Beautiful Relation
$$v_p \times v_g = c^2$$Interactive Demo: de Broglie Wavelength Calculator
const DeBroglieCalculator = () => {
const [particle, setParticle] = useState('electron');
const [velocity, setVelocity] = useState(1e6); // m/s
const particles = {
electron: { mass: 9.11e-31, charge: 1.6e-19 },
proton: { mass: 1.67e-27, charge: 1.6e-19 },
neutron: { mass: 1.67e-27, charge: 0 },
ball: { mass: 0.05, charge: 0 } // 50g ball
};
const h = 6.626e-34;
const m = particles[particle].mass;
const lambda = h / (m * velocity);
return (
<div>
<h3>de Broglie Wavelength Calculator</h3>
<label>Particle:</label>
<select value={particle} onChange={(e) => setParticle(e.target.value)}>
<option value="electron">Electron</option>
<option value="proton">Proton</option>
<option value="neutron">Neutron</option>
<option value="ball">Cricket Ball (50g)</option>
</select>
<label>Velocity: {velocity.toExponential(2)} m/s</label>
<input
type="range"
min="1e3"
max="1e7"
value={velocity}
onChange={(e) => setVelocity(Number(e.target.value))}
/>
<div className="result">
<h4>Results:</h4>
<p>Mass: {m.toExponential(2)} kg</p>
<p>Momentum: {(m * velocity).toExponential(2)} kg·m/s</p>
<p>de Broglie wavelength: {lambda.toExponential(2)} m</p>
{lambda > 1e-10 ? (
<p style={{color: 'green'}}>
✓ Observable wave nature! ({(lambda * 1e10).toFixed(2)} Å)
</p>
) : (
<p style={{color: 'red'}}>
Wave nature negligible for macroscopic objects
</p>
)}
<p>For comparison:</p>
<ul>
<li>Atomic spacing: ~1-3 Å</li>
<li>Visible light: 4000-7000 Å</li>
<li>X-rays: 0.1-10 Å</li>
</ul>
</div>
</div>
);
};
Why Don’t We See Cricket Balls Diffract?
Example Calculation
Cricket ball: m = 0.05 kg, v = 40 m/s
$$λ = \frac{h}{mv} = \frac{6.626 × 10^{-34}}{0.05 × 40} = 3.3 × 10^{-34} \text{ m}$$This is impossibly small! (10²⁴ times smaller than an atom)
The Rule of Thumb
Wave nature is significant when:
$$λ \sim \text{size of obstacle/slit}$$- Electrons: λ ~ 1 Å, atomic spacing ~ 1 Å → Diffraction visible!
- Macroscopic objects: λ ~ 10⁻³⁴ m → No diffraction
Bohr’s Quantization from de Broglie
The Circular Standing Wave Condition
For an electron in a stable orbit around nucleus:
Orbit
___________
/ λ \
| ← → ← → ← | Standing wave
\___________/
Circumference = nλ
Substituting λ = h/mv:
$$2πr = n\frac{h}{mv}$$ $$mvr = \frac{nh}{2π} = n\hbar$$This is Bohr’s quantization condition! 🎯
de Broglie explained WHY angular momentum is quantized - electrons are standing waves!
Memory Tricks 🧠
“LAMBDA PIZZA” for de Broglie
Light has λ = c/ν All particles have waves Momentum determines λ Big mass → small λ De Broglie: λ = h/p Accelerated charges: use qV
Planck’s h is the bridge Inverse relation with p Zero momentum → infinite λ Zig-zag electron paths are waves Angular momentum quantized
The MASS-SPEED Memory
Heavy & Slow → LARGE λ → Wave-like
Light & Fast → SMALL λ → Particle-like
Wait, that's backwards! 🤔
Actually:
Heavy OR Slow → small λ (mv in denominator)
Light AND Fast → can have large λ
Electron (light) at normal speeds → λ ~ Å ✓
Bullet (heavy) at any speed → λ ~ 10⁻³⁴ m
Quick Conversion Formulas
For electrons accelerated through V volts:
$$λ(\text{Å}) = \frac{12.27}{\sqrt{V}}$$For kinetic energy K in eV:
$$λ(\text{Å}) = \frac{12.27}{\sqrt{K}}$$Common Mistakes ⚠️
❌ Mistake 1: Using wrong mass
Wrong: Using rest mass for relativistic particles Right: Use relativistic momentum p = γm₀v for v ~ c
❌ Mistake 2: Confusing λ with wavelength of light
Wrong: “Electron absorbs light of wavelength λ” Right: Electron HAS associated wavelength λ (not light!)
❌ Mistake 3: Forgetting √2m
Wrong: λ = h/√(mK) Right: λ = h/√(2mK)
❌ Mistake 4: Mixing up phase and group velocity
Wrong: “Particle travels at vₚ = c²/v” Right: Particle travels at vₘ = v (group velocity)
❌ Mistake 5: Wrong units
Wrong: V in mV, getting λ in wrong units Right: V must be in volts for λ(Å) = 12.27/√V
Important Comparisons
de Broglie Wavelengths at Room Temperature (300 K)
Using thermal energy: K = (3/2)kT = 0.039 eV
| Particle | Mass (kg) | λ (Å) |
|---|---|---|
| Electron | 9.11 × 10⁻³¹ | 62 |
| Proton | 1.67 × 10⁻²⁷ | 1.4 |
| H₂ molecule | 3.34 × 10⁻²⁷ | 1.0 |
| He atom | 6.68 × 10⁻²⁷ | 0.7 |
Electron Wavelengths at Different Energies
| Accelerating Voltage | K (eV) | λ (Å) |
|---|---|---|
| 10 V | 10 | 3.88 |
| 100 V | 100 | 1.23 |
| 1000 V | 1 keV | 0.388 |
| 10,000 V | 10 keV | 0.123 |
Note: At 10 keV, relativistic effects ~1%
Derivations You Must Know
1. λ from Kinetic Energy
$$K = \frac{1}{2}mv^2$$ $$p^2 = 2mK$$ $$p = \sqrt{2mK}$$ $$λ = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$$2. λ for Accelerated Electron
$$K = eV$$ $$λ = \frac{h}{\sqrt{2meV}}$$Substituting values:
- h = 6.626 × 10⁻³⁴ J·s
- m = 9.11 × 10⁻³¹ kg
- e = 1.6 × 10⁻¹⁹ C
3. Bohr Radius from de Broglie
For ground state (n = 1):
$$2πr_1 = λ$$ $$2πr_1 = \frac{h}{mv}$$Also, for circular motion:
$$\frac{mv^2}{r_1} = \frac{ke^2}{r_1^2}$$Combining these:
$$r_1 = \frac{h^2}{4π^2mke^2} = 0.529 \text{ Å}$$This is the Bohr radius!
Problem-Solving Strategy
Step 1: Identify Given Information
- Mass of particle
- Velocity OR kinetic energy OR potential
- Sometimes wavelength → find other quantities
Step 2: Choose Appropriate Formula
Known? Use Formula
------ -----------
v λ = h/mv
K λ = h/√(2mK)
V (potential) λ = h/√(2mqV)
E (total) Find K first, then use above
Step 3: Unit Consistency
| Quantity | SI Unit | Alternative |
|---|---|---|
| h | 6.626 × 10⁻³⁴ J·s | 4.14 × 10⁻¹⁵ eV·s |
| m | kg | - |
| v | m/s | - |
| K | J | eV (1.6 × 10⁻¹⁹ J) |
| λ | m | Å (10⁻¹⁰ m) |
Practice Problems
Level 1: JEE Main Basics
Q1. An electron is accelerated through 100 V. Find its de Broglie wavelength.
Solution:
Using λ = 12.27/√V Å
λ = 12.27/√100 = 12.27/10 = 1.227 Å
Q2. A particle of mass 10⁻⁶ kg is moving with velocity 10⁻² m/s. Calculate its de Broglie wavelength.
Solution:
λ = h/mv
λ = (6.626 × 10⁻³⁴)/(10⁻⁶ × 10⁻²)
λ = 6.626 × 10⁻²⁶ m
This is extremely small - wave nature negligible!
Q3. An electron and a proton have the same kinetic energy. Find the ratio of their de Broglie wavelengths.
Solution:
λ = h/√(2mK)
λₑ/λₚ = √(mₚ/mₑ) = √(1836) ≈ 43
Electron wavelength is 43 times larger!
Level 2: JEE Main/Advanced
Q4. An electron has de Broglie wavelength equal to the wavelength of yellow light (600 nm). Find: (a) Momentum of electron (b) Kinetic energy in eV (c) Accelerating potential needed
Solution:
(a) p = h/λ = (6.626 × 10⁻³⁴)/(600 × 10⁻⁹)
p = 1.104 × 10⁻²⁷ kg·m/s
(b) K = p²/2m = (1.104 × 10⁻²⁷)²/(2 × 9.11 × 10⁻³¹)
K = 6.69 × 10⁻²⁵ J = 4.18 × 10⁻⁶ eV
K ≈ 4.18 μeV (very small!)
(c) V = K/e = 4.18 × 10⁻⁶ V ≈ 4.18 μV
Q5. A proton and an α-particle are accelerated through the same potential difference. Find the ratio of their de Broglie wavelengths.
Solution:
λ = h/√(2mqV)
For same V:
λ ∝ 1/√(mq)
Proton: m = mₚ, q = e
α-particle: m = 4mₚ, q = 2e
λₚ/λₐ = √(mₐqₐ)/(mₚqₚ) = √(4mₚ × 2e)/(mₚ × e)
λₚ/λₐ = √8 = 2√2 ≈ 2.83
Q6. Show that for a photon, de Broglie wavelength equals electromagnetic wavelength.
Solution:
For photon:
E = pc (massless)
E = hν = hc/λ_EM
p = E/c = h/λ_EM
de Broglie: λ_dB = h/p = h/(h/λ_EM) = λ_EM
Perfect match! ✓
Level 3: JEE Advanced
Q7. An electron in hydrogen atom jumps from n=3 to n=1 orbit. The emitted photon strikes a metal surface with work function 2 eV. Find: (a) de Broglie wavelength of emitted photon (b) de Broglie wavelength of ejected photoelectron (Energy levels: Eₙ = -13.6/n² eV)
Solution:
(a) E₃ = -13.6/9 = -1.51 eV
E₁ = -13.6 eV
Photon energy: hν = E₃ - E₁ = -1.51 - (-13.6) = 12.09 eV
λ_photon = hc/E = 1240 eV·nm / 12.09 eV = 102.6 nm
(b) K_max = hν - φ = 12.09 - 2 = 10.09 eV
For electron:
λ = 12.27/√K Å = 12.27/√10.09 = 3.86 Å
Q8. An electron and a photon have the same wavelength of 0.1 nm. Compare their: (a) Momentum (b) Energy
Solution:
(a) Both have same momentum!
p = h/λ = (6.626 × 10⁻³⁴)/(0.1 × 10⁻⁹)
p = 6.626 × 10⁻²⁴ kg·m/s
(b) For photon:
E = pc = 6.626 × 10⁻²⁴ × 3 × 10⁸
E = 1.988 × 10⁻¹⁵ J = 12.4 keV
For electron:
E = p²/2m = (6.626 × 10⁻²⁴)²/(2 × 9.11 × 10⁻³¹)
E = 2.41 × 10⁻¹⁷ J = 150 eV
E_photon/E_electron = 12400/150 ≈ 83
Photon has 83 times more energy for same wavelength!
Q9. An electron is confined in a one-dimensional box of length L. Using de Broglie hypothesis, show that allowed energies are quantized.
Solution:
For standing wave in box: L = n(λ/2)
where n = 1, 2, 3, ...
λ = 2L/n
Using de Broglie: λ = h/p
2L/n = h/p
p = nh/2L
Energy: E = p²/2m = (nh/2L)²/2m
E = n²h²/8mL²
This is the particle-in-a-box quantization!
Allowed energies: E₁, E₂, E₃, ... ∝ n²
Q10. A neutron beam is used for diffraction from crystal planes with spacing d = 2 Å. If the first-order maximum occurs at θ = 30°, find: (a) de Broglie wavelength (b) Velocity of neutrons (c) Temperature corresponding to this thermal energy
Solution:
(a) Bragg's law: 2d sinθ = nλ
For n = 1: λ = 2d sinθ = 2 × 2 × sin30°
λ = 2 × 2 × 0.5 = 2 Å
(b) λ = h/mv
v = h/mλ
v = (6.626 × 10⁻³⁴)/(1.67 × 10⁻²⁷ × 2 × 10⁻¹⁰)
v = 1.98 × 10³ m/s
(c) K = ½mv² = ½ × 1.67 × 10⁻²⁷ × (1.98 × 10³)²
K = 3.28 × 10⁻²¹ J
K = (3/2)kT
T = 2K/3k = (2 × 3.28 × 10⁻²¹)/(3 × 1.38 × 10⁻²³)
T = 158 K ≈ -115°C
This is cryogenic temperature!
Cross-Links to Related Topics
- Photoelectric Effect - Wave-particle duality of light
- Davisson-Germer Experiment - Experimental proof of de Broglie hypothesis
- Bohr Model - Angular momentum quantization explained
- Hydrogen Spectrum - Energy levels and transitions
Quick Revision Checklist ✓
- de Broglie relation: λ = h/p = h/mv
- For kinetic energy: λ = h/√(2mK)
- For accelerated electron: λ = 12.27/√V Å
- Wave-particle duality applies to ALL matter
- Massive/fast particles → small λ → particle-like
- Light/slow particles → large λ → wave-like
- Bohr quantization: 2πr = nλ
- Phase velocity: vₚ = c²/v
- Group velocity: vₘ = v (particle velocity)
- Photon: λ_dB = λ_EM (same!)
Conceptual Questions for Deep Understanding
Q1. Why can’t we observe diffraction of cricket balls? A: λ ~ 10⁻³⁴ m, far smaller than any obstacle. Need λ ~ size for diffraction.
Q2. Can a photon have a de Broglie wavelength? A: Yes! λ = h/p, and for photon p = E/c, so λ = hc/E = electromagnetic λ.
Q3. Why does heavier particle have smaller wavelength at same speed? A: λ = h/mv. Larger m (mass) in denominator → smaller λ.
Q4. Can macroscopic objects behave like waves? A: In principle yes, but λ is so tiny it’s unobservable. Quantum effects vanish for large masses.
Q5. What happens to λ as particle approaches speed of light? A: Must use relativistic momentum. λ → 0 as v → c (for massive particles).
Final Tips for JEE
- Master the formula: λ = h/p in all forms
- Quick electron formula: λ(Å) = 12.27/√V - memorize this!
- Proportionalities: λ ∝ 1/v, λ ∝ 1/√K, λ ∝ 1/√V
- Same K: λ ∝ 1/√m (lighter particle → larger λ)
- Same V: λ ∝ 1/√(mq) (compare different particles)
- Units: Keep h in J·s, get λ in m; OR use shortcuts
- Bohr connection: Standing waves → quantization
- Numerical values: Know electron/proton mass ratio = 1836
Last updated: April 18, 2025 Previous: Photoelectric Effect Next: Davisson-Germer Experiment