Q.The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10−15m or less. This structure was first probed in early 1970's using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511MeV.)
De Broglie Wavelength: When Particles Start Acting Like Waves
You already know that light behaves like a wave (interference, diffraction) and like a particle (photoelectric effect). That's wave-particle duality for light. De Broglie's radical idea in 1924 was: if light can be both, why can't matter be both too?
He proposed that every moving particle — an electron, a proton, even a cricket ball — has a wavelength associated with it. The faster it moves, the shorter that wavelength becomes.
The Intuition
Think of a wave on a string. Its wavelength is the distance between two consecutive crests. Now imagine an electron moving through space. De Broglie said that the electron's motion itself creates a "matter wave" — a wave of probability that guides where the electron is likely to be found.
Note
You never see this wavelength in everyday life because for large objects it's unimaginably tiny. A cricket ball moving at 30 m/s has a de Broglie wavelength of about 10−34 m — far smaller than an atomic nucleus. That's why macroscopic objects behave like particles.
The Precise Statement
The de Broglie wavelength λ of a particle is given by:
λ=ph
where:
h is Planck's constant (6.626×10−34J⋅s)
p is the momentum of the particle (p=mv for non-relativistic speeds)
Key point: The wavelength depends only on momentum, not on charge, mass, or any other property. A fast electron and a slow proton can have the same wavelength if their momenta are equal.
What This Means Physically
For electrons in atoms: The de Broglie wavelength of an electron in a hydrogen atom is roughly the size of the atom itself (≈10−10 m). This is why electrons form standing waves around the nucleus — only certain wavelengths "fit" into the orbit, which explains quantised energy levels.
For experiments: If you fire electrons through a crystal, they diffract just like X-rays. This was confirmed by Davisson and Germer in 1927 — a Nobel-winning experiment that proved de Broglie right.
For large objects: The wavelength is so small that wave behaviour is undetectable. A car moving at 100 km/h has λ≈10−38 m — you'd need a slit smaller than an atom to see diffraction.
Watch out
A common mistake is to think the de Broglie wavelength is the size of the particle. It is not. It is the wavelength of the probability wave associated with the particle. The particle itself remains point-like.
Worked Example
Question: What is the de Broglie wavelength of an electron moving at 2.0×106 m/s? (Mass of electron me=9.11×10−31 kg)
Solution:
First, find momentum:
p=mv=(9.11×10−31)(2.0×106)=1.822×10−24kg⋅m/s
Then apply de Broglie's formula: …
Why this formula?
De Broglie Wavelength: Why Matter Has a Wave Nature
The idea that a moving particle has a wavelength is one of the most radical shifts in physics. It came from Louis de Broglie in 1924, who asked a simple question: if light — which we thought was a wave — can behave like a particle (the photon), then why can't a particle behave like a wave?
The Core Insight: Symmetry in Nature
De Broglie started from Einstein's relation for a photon. For light, the energy E and momentum p of a photon are linked to its wave properties — frequency f and wavelength λ — by:
E=hfandp=λh
where h is Planck's constant. These are not arbitrary; they come from the fact that light is an electromagnetic wave, and Planck had already shown that energy comes in quanta hf.
De Broglie's reasoning was a leap of symmetry: if nature treats light and matter on equal footing (as Einstein's special relativity suggests), then any moving particle should also have a wavelength associated with it. He proposed that the same relation holds for matter:
λ=ph
where p=mv is the momentum of the particle (for non-relativistic speeds). This is the de Broglie wavelength.
Why This Formula Makes Sense: A Simple Derivation
There is no rigorous "derivation" from first principles — de Broglie's hypothesis was a postulate. But we can see why it is plausible by combining two key ideas from relativity and quantum theory.
Step 1: Energy of a particle from relativity
For a particle with rest mass m0, the total energy in special relativity is:
E=p2c2+m02c4
For a photon, m0=0, so E=pc. This matches the photon's wave relation E=hf and p=h/λ.
Step 2: Assume the same wave-particle duality for matter
If a massive particle also has a wave associated with it, then its energy should also be E=hf, where f is the frequency of the matter wave. Equating the relativistic energy with the quantum energy:
hf=p2c2+m02c4
For a particle moving at non-relativistic speeds (v≪c), the momentum p=mv is small compared to m0c, so we can expand:
E≈m0c2+2m0p2
The first term is rest energy, which is constant. The second term is kinetic energy K=p2/(2m). The wave frequency f then corresponds to the kinetic part (since rest energy doesn't contribute to motion). But the key relation we want is between wavelength and momentum.
Step 3: The wavelength from the wave speed
For any wave, the speed vwave=fλ. For a matter wave, de Broglie proposed that the wave speed equals the particle's speed v (this is the phase velocity). So:
v=fλ
Now use E=hf and E=21mv2 (non-relativistic kinetic energy). Then:
f=hE=2hmv2
Substitute into v=fλ:
v=2hmv2λ⇒λ=mv2h
This gives λ=2h/p, which is wrong by a factor of 2. The correct formula is λ=h/p. …
To probe a structure as small as 10−15m, the electron beam's de Broglie wavelength must be of that same order, which via λ=h/p demands a momentum — and hence, since the electron is then moving essentially at the speed of light, an energy — …
A probing wavelength of 10−15 m requires momentum p=h/λ≈6.6×10−19 kg m/s. Since pc then works out to be over 1000 times the electron's rest mass energy, the electron is ultra-relativistic and E≈pc≈1.24 GeV, of order 1 GeV.
Step 1 — Momentum needed for λ∼10−15m.
p=λh=10−156.63×10−34=6.63×10−19kg m/s
Step 2 — Check whether the electron is relativistic.
Comparing to the electron's rest mass energy, mec2=0.511MeV=0.000511GeV: since pc≫mec2 by more than a factor of 2000, the electron must be ultra-relativistic.