Dual nature: light first, then matter
Light shows interference and diffraction, so it is a wave. In the photoelectric effect it gives energy in packets, so it is a particle (photon). Light has a dual nature.
Louis de Broglie asked a simple question in 1924: nature likes balance, so if waves can act like particles, can particles act like waves? He said yes. Every moving piece of matter has a wave linked to it. We call it a matter wave or de Broglie wave.
The de Broglie relation
For a photon, momentum p = h/λ. De Broglie used the same rule for matter:
λ = h / p = h / (m v)
Here h = 6.63 × 10⁻³⁴ J s, m is the mass and v the speed of the particle.
- Bigger momentum → shorter wavelength.
- A particle at rest (p = 0) has no matter wave (λ would be infinite).
- Matter waves are not electromagnetic waves; they are found with charged and uncharged particles alike.
λ in terms of kinetic energy and voltage
Kinetic energy K = ½mv² = p²/2m, so p = √(2mK) and
λ = h / √(2mK)
A particle of charge q accelerated from rest through a potential difference V gets K = qV:
λ = h / √(2mqV)
For an electron (m = 9.11 × 10⁻³¹ kg, q = 1.6 × 10⁻¹⁹ C) this becomes
λ = 1.227 / √V nm (V in volts).
At V = 100 V, λ ≈ 0.123 nm. That is similar to the gap between atoms in a crystal, which is why crystals can diffract electrons.
Why don't we see the wave of a ball?
A 0.16 kg cricket ball at 36 m/s has p ≈ 5.8 kg m/s, so λ ≈ 1.1 × 10⁻³⁴ m. That is far smaller than even a nucleus (about 10⁻¹⁵ m). No slit or crystal is that small, so its wave effects can never be seen. Waves show up only when λ is comparable to the size of the obstacle. So wave nature matters for tiny particles like electrons and neutrons, not for everyday objects.
Comparing particles and a photon
- Same kinetic energy: λ ∝ 1/√m. An electron has a longer wavelength than a proton with the same K.
- Same voltage: λ ∝ 1/√(mq). Electron > proton > alpha particle.
- Same speed: λ ∝ 1/m.
- Same momentum: same λ, whatever the particle.
- A photon's energy E = pc, while a particle's K = p²/2m. So for the same wavelength, a photon and an electron have different energies.
Try it: predict, then check
- In the 3D pick Electron and set V = 100. Note λ. Predict λ at V = 400, then check (it should halve).
- Keep V = 100. Switch to Proton, then Alpha particle. Order the three wavelengths before you look.
- Pick Cricket ball. Move the slider from 1 to 60 m/s. Does the line ever wiggle? Why not?
- At home: throw a ball through a doorway. It never spreads out like sound does. Compare your λ (mass about 50 kg, walking at 1 m/s) with a door width.
Key formulas and definitions
- λ = h/p = h/(mv)
- λ = h/√(2mK)
- λ = h/√(2mqV)
- Electron: λ = 1.227/√V nm
- Photon: p = h/λ = E/c
- Same K: λ ∝ 1/√m; same V: λ ∝ 1/√(mq)
Worked examples
1. Find the de Broglie wavelength of a 0.16 kg ball moving at 36 m/s.
Step 1: p = mv = 0.16 × 36 = 5.76 kg m/s. Step 2: λ = h/p = 6.63 × 10⁻³⁴ / 5.76. Step 3: λ ≈ 1.15 × 10⁻³⁴ m, far too small to observe.
2. An electron moves at 2 × 10⁶ m/s. Find its de Broglie wavelength.
Step 1: p = mv = 9.11 × 10⁻³¹ × 2 × 10⁶ = 1.822 × 10⁻²⁴ kg m/s. Step 2: λ = 6.63 × 10⁻³⁴ / 1.822 × 10⁻²⁴. Step 3: λ ≈ 3.64 × 10⁻¹⁰ m = 0.364 nm.
3. An electron is accelerated through 144 V. Find its de Broglie wavelength.
Step 1: Use λ = 1.227/√V nm. Step 2: √144 = 12. Step 3: λ = 1.227/12 ≈ 0.102 nm.
4. Find the wavelength of an electron with kinetic energy 50 eV.
Step 1: K = 50 × 1.6 × 10⁻¹⁹ = 8 × 10⁻¹⁸ J. Step 2: p = √(2mK) = √(2 × 9.11 × 10⁻³¹ × 8 × 10⁻¹⁸) = √(1.458 × 10⁻⁴⁷) ≈ 3.82 × 10⁻²⁴ kg m/s. Step 3: λ = h/p = 6.63 × 10⁻³⁴ / 3.82 × 10⁻²⁴ ≈ 1.74 × 10⁻¹⁰ m = 0.174 nm. (Check: 1.227/√50 = 0.174 nm.)
5. A proton and an electron have the same kinetic energy. Find the ratio λe : λp (mp = 1836 me).
Step 1: λ = h/√(2mK); with K the same, λ ∝ 1/√m. Step 2: λe/λp = √(mp/me) = √1836. Step 3: λe : λp ≈ 42.8 : 1.
6. A proton and an alpha particle are accelerated through the same voltage. Find λp : λα (mα = 4mp, qα = 2qp).
Step 1: λ = h/√(2mqV) → λ ∝ 1/√(mq). Step 2: λp/λα = √(mα qα / mp qp) = √(4 × 2) = √8. Step 3: λp : λα = 2√2 : 1 ≈ 2.83 : 1.
7. Through what voltage must an electron be accelerated to have a wavelength of 0.1 nm?
Step 1: λ = 1.227/√V → √V = 1.227/λ = 1.227/0.1 = 12.27. Step 2: V = 12.27² ≈ 150.6 V. Step 3: So about 151 V.
Common mistakes
- Using the formula 1.227/√V nm for protons or alpha particles. It works only for electrons.
- Forgetting to change eV into joules before using λ = h/√(2mK).
- Thinking matter waves are electromagnetic waves. They are not; even neutral particles like neutrons have them.
- Saying a heavier particle always has a shorter wavelength. At the same momentum the wavelength is the same whatever the mass.