Q.Assuming an electron is confined to a $1\ \text{nm}$ wide region, find the uncertainty in momentum using the Heisenberg Uncertainty principle. You can assume the uncertainty in position $\Delta x$ as $1\ \text{nm}$. Assuming $p \approx \Delta p$, find the energy of the electron in electron volts.

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 $\lambda$ — by:

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:

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 $m_0$, the total energy in special relativity is:

For a photon, $m_0 = 0$, so $E = pc$. This matches the photon's wave relation $E = hf$ and $p = h/\lambda$.

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:

For a particle moving at non-relativistic speeds ($v \ll c$), the momentum $p = mv$ is small compared to $m_0 c$, so we can expand:

The first term is rest energy, which is constant. The second term is kinetic energy $K = p^2/(2m)$. 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 $v_{\text{wave}} = f \lambda$. For a matter wave, de Broglie proposed that the wave speed equals the particle's speed $v$ (this is the phase velocity). So:

Now use $E = hf$ and $E = \frac{1}{2}mv^2$ (non-relativistic kinetic energy). Then:

Substitute into $v = f \lambda$:

This gives $\lambda = 2h/p$, which is wrong by a factor of 2. The correct formula is $\lambda = h/p$. …