de Broglie Wavelength — Matter Waves Calculated
Light had been shown to behave as both a wave and a stream of particles. Louis de Broglie's proposal, in 1924, turned that around: if a wave can act like a particle, a particle must be able to act like a wave. Every moving object — an electron, a neutron, a cricket ball — has a wavelength. The formula is one line; what makes it worth understanding is what happens when you put real numbers into it, because the answers explain both why electrons behave strangely and why cricket balls do not.
The relation
| Symbol | Meaning | SI unit |
|---|---|---|
| λ | de Broglie wavelength of the moving particle | metre, m |
| h | Planck constant, 6.626 × 10⁻³⁴ (defined exactly as 6.62607015 × 10⁻³⁴) | J·s |
| p | Linear momentum of the particle | kg·m·s⁻¹ |
| m | Mass — for an electron, 9.109 × 10⁻³¹ kg | kilogram, kg |
| v | Speed | m·s⁻¹ |
Two rearranged forms save time in exams. If a particle has kinetic energy K, then p = √(2mK), so
and for a charge q accelerated from rest through a potential difference V, where K = qV:
λ = h / √(2 m q V)
For an electron the second form collapses into a very useful shortcut, which is derived in worked example 2:
Worked example 1 — an electron at 10⁶ m/s
Question: Find the de Broglie wavelength of an electron moving at 1.0 × 10⁶ m s⁻¹.
Step 1 — momentum.
p = m v = (9.109 × 10⁻³¹ kg)(1.0 × 10⁶ m s⁻¹) = 9.109 × 10⁻²⁵ kg·m·s⁻¹
Step 2 — wavelength.
λ = h / p = (6.626 × 10⁻³⁴) / (9.109 × 10⁻²⁵)
Divide the numbers: 6.626 ÷ 9.109 = 0.7274. Subtract the powers: 10⁻³⁴ ÷ 10⁻²⁵ = 10⁻⁹.
λ = 0.7274 × 10⁻⁹ = 7.27 × 10⁻¹⁰ m = 0.727 nm = 7.27 Å
Why this number matters: atoms in a crystal are spaced a few times 10⁻¹⁰ m apart, the same size as this wavelength — so a crystal can act as a diffraction grating for the electron.
Worked example 2 — an electron accelerated through 100 V
Question: An electron starts from rest and is accelerated through a potential difference of 100 V. Find its de Broglie wavelength.
Step 1 — kinetic energy gained.
K = qV = (1.602 × 10⁻¹⁹ C)(100 V) = 1.602 × 10⁻¹⁷ J
Step 2 — momentum from the energy.
2mK = 2 × (9.109 × 10⁻³¹) × (1.602 × 10⁻¹⁷) = 2.9185 × 10⁻⁴⁷
p = √(2.9185 × 10⁻⁴⁷). Rewrite as 29.185 × 10⁻⁴⁸ so the power is even:
p = √29.185 × 10⁻²⁴ = 5.402 × 10⁻²⁴ kg·m·s⁻¹
Step 3 — wavelength. λ = (6.626 × 10⁻³⁴)/(5.402 × 10⁻²⁴) = 1.2265 × 10⁻¹⁰ m = 0.1227 nm = 1.227 Å
Cross-check with the shortcut: λ = 1.227/√100 = 0.1227 nm ✔ — the shortcut is just this calculation done once with the constants left as symbols.
Is the non-relativistic formula safe here? v = p/m = 5.93 × 10⁶ m s⁻¹, which is 1.98% of the speed of light — so yes. Above roughly 10 kV the relativistic correction starts to matter and the simple formula over-estimates λ.
Worked example 3 — a cricket ball
Question: A cricket ball of mass 0.16 kg is bowled at 40 m s⁻¹ (144 km/h). Find its de Broglie wavelength.
Step 1 — momentum. p = 0.16 × 40 = 6.4 kg·m·s⁻¹
Step 2 — wavelength.
λ = (6.626 × 10⁻³⁴)/6.4 = 1.04 × 10⁻³⁴ m
Put that in perspective. An atomic nucleus is about 10⁻¹⁵ m across, so this wavelength is roughly 10⁻¹⁹ times smaller than a nucleus. A walking person of mass 60 kg at 1.5 m s⁻¹ does even worse: p = 90 kg·m·s⁻¹ and λ = 6.626 × 10⁻³⁴/90 = 7.4 × 10⁻³⁶ m.
Why the wave nature is invisible at everyday scale
Wave behaviour — diffraction and interference — only shows itself when the wavelength is comparable to the slit, obstacle or spacing the wave meets. Sound bends round a doorway because its wavelength is about a metre; light does not, because its wavelength is under a micrometre.
A cricket ball's wavelength is 10⁻³⁴ m, and no aperture is small enough to diffract it: even a slit of 10⁻⁶ m would spread the ball by an angle of order λ/d ≈ 10⁻²⁸ radian, far below anything detectable. Turn the question around and it is clearer still. What speed would a cricket ball need for its wavelength to reach an atomic 10⁻¹⁰ m?
At that speed the ball would take about 2.4 × 10²² seconds — roughly 8 × 10¹⁴ years, more than fifty thousand times the present age of the universe — to travel one metre. So the wave nature of ordinary objects is not switched off; it is pushed so far into the unmeasurable that it can never show up. The reason is that h is minute (10⁻³⁴) while everyday masses are not. Only when mv falls to around 10⁻²⁴ does λ climb to atomic size, and that needs a particle as light as an electron.
The evidence, and the link back to Bohr
The idea was confirmed experimentally in 1927, when Clinton Davisson and Lester Germer observed electrons scattered from a nickel crystal producing a diffraction pattern of the kind only a wave can make; G. P. Thomson independently obtained diffraction rings from electrons passing through thin films. The same principle underlies electron microscopy: since resolution is limited by wavelength, and an accelerated electron's wavelength is thousands of times shorter than visible light, an electron microscope resolves detail an optical one never could.
There is also a neat link to the Bohr model. Bohr had to assume that angular momentum is quantised as mvr = nh/2π. De Broglie's picture explains it: a stable orbit is one whose circumference holds a whole number of electron wavelengths, 2πr = nλ. Check it for the ground state of hydrogen, where r = 5.29 × 10⁻¹¹ m and the electron speed is about 2.188 × 10⁶ m s⁻¹:
- p = (9.109 × 10⁻³¹)(2.188 × 10⁶) = 1.993 × 10⁻²⁴ kg·m·s⁻¹, so λ = 6.626 × 10⁻³⁴/1.993 × 10⁻²⁴ = 3.325 × 10⁻¹⁰ m
- 2πr = 2 × 3.1416 × 5.29 × 10⁻¹¹ = 3.324 × 10⁻¹⁰ m
The two agree to about 0.04% — exactly one wavelength fits the first orbit, as n = 1 requires.
Worked example 4 — the photon, where it all started
Question: Find the energy and momentum of a photon of green light, λ = 500 nm. Take c = 3.00 × 10⁸ m s⁻¹.
Energy: E = hc/λ, with hc = (6.626 × 10⁻³⁴)(3.00 × 10⁸)
= 1.9878 × 10⁻²⁵ J·m and λ = 5.00 × 10⁻⁷ m
E = (1.9878 × 10⁻²⁵)/(5.00 × 10⁻⁷) = 3.976 × 10⁻¹⁹ J
In electronvolts: 3.976 × 10⁻¹⁹ / 1.602 × 10⁻¹⁹ = 2.48 eV
Momentum: a photon has no rest mass, so p = mv cannot be used — but
λ = h/p still holds:
p = h/λ = (6.626 × 10⁻³⁴)/(5.00 × 10⁻⁷) = 1.33 × 10⁻²⁷ kg·m·s⁻¹
The electron of example 2 had momentum 5.40 × 10⁻²⁴ — about 4000 times larger, which is exactly why its wavelength is about 4000 times shorter.
Common mistakes that cost marks
- Using mass in grams. The electron mass is 9.109 × 10⁻³¹ kg, not grams. A factor of 1000 in the mass appears directly in the answer.
- Leaving the kinetic energy in electronvolts. K must be in joules before it enters λ = h/√(2mK). Multiply eV by 1.602 × 10⁻¹⁹ first.
- Forgetting the square root. λ = h/√(2mK), not h/(2mK). Dropping the root changes the order of magnitude completely.
- Writing λ = h/(mv) for a photon. A photon has zero rest mass; use p = h/λ = E/c instead.
- Thinking a matter wave is an electromagnetic wave. It is not. It describes where the particle is likely to be found — the amplitude relates to probability, not to an electric or magnetic field.
- Sloppy powers of ten. Split the working as in example 1 — divide the decimal parts first, subtract the exponents second. Most errors here are arithmetic, not conceptual.
Where this appears in exams
| Exam / subject | Typical use |
|---|---|
| CBSE/ICSE Class 12 Physics | Dual Nature of Radiation and Matter — de Broglie wavelength, photoelectric effect, accelerating-voltage numericals |
| Class 11 Chemistry | Structure of Atom — de Broglie relation, Heisenberg uncertainty, why orbits become orbitals |
| JEE Main & Advanced | Combined energy–momentum–wavelength problems, comparisons between particles |
| BSc / IIT-JAM level | Wave–particle duality as the entry point to the Schrödinger equation |
Check your duality numericals instantly. Enter a mass and speed, a kinetic energy, an accelerating voltage or a photon wavelength, and the Modern Physics tool returns the matching wavelength, momentum and energy — so a stray power of ten is caught before it reaches your answer sheet.
Open the Modern Physics (de Broglie & Photon) Calculator →Class 12 physics and chemistry both cover this topic, from opposite directions. ABC Chemistry runs Class 11–12 coaching at the Gurugram centre plus online classes across India — details at abcchemistry.in.