Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature ($27\ °\text{C}$) and $1\ \text{atm}$ pressure; and compare it with the mean separation between two atoms under these conditions.

Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27°C) and

Additional Exercises · 11.35

27\ °\text{C}

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Concept understanding — De Broglie Wavelength

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 $\lambda$ of a particle is given by:

\lambda = \frac{h}{p}

where:

$h$ is Planck's constant ( $6.626 \times 10^{-34} \ \text{J·s}$ )
$p$ 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 ( $\approx 10^{-10}$ 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 $\lambda \approx 10^{-38}$ 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 $2.0 \times 10^6$ m/s? (Mass of electron $m_e = 9.11 \times 10^{-31}$ kg)

Solution:

First, find momentum:

p = mv = (9.11 \times 10^{-31})(2.0 \times 10^6) = 1.822 \times 10^{-24} \ \text{kg·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 $\lambda$ — by:

E = h f \quad \text{and} \quad p = \frac{h}{\lambda}

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:

\lambda = \frac{h}{p}

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:

E = \sqrt{p^2 c^2 + m_0^2 c^4}

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:

hf = \sqrt{p^2 c^2 + m_0^2 c^4}

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:

E \approx m_0 c^2 + \frac{p^2}{2m_0}

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:

v = f \lambda

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

f = \frac{E}{h} = \frac{mv^2}{2h}

Substitute into $v = f \lambda$ :

v = \frac{mv^2}{2h} \lambda \quad \Rightarrow \quad \lambda = \frac{2h}{mv}

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

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Using the rms-based momentum $p=\sqrt{3mkT}$ for a He atom at 300 K gives $\lambda\approx0.73$ Å, while the ideal gas law gives a mean atomic separation of ≈34.5 Å — about 47 times larger, so helium atoms in a gas do not have overlapping wave packets and behave as classically distinguishable particles.

Step 1 — de Broglie wavelength of a He atom.

Mass of a He atom: $m = 4.0026\ \text{u} \times 1.66\times10^{-27} \approx 6.644\times10^{-27}\ \text{kg}$

p = \sqrt{3mkT} = \sqrt{3(6.644\times10^{-27})(1.38\times10^{-23})(300)}

= \sqrt{3(6.644\times10^{-27})(4.14\times10^{-21})} = \sqrt{8.25\times10^{-47}} \approx 9.08\times10^{-24}\ \text{kg m/s}

\lambda = \frac{h}{p} = \frac{6.63\times10^{-34}}{9.08\times10^{-24}} \approx 7.30\times10^{-11}\ \text{m} = 0.73\ \text{Å}

Step 2 — Mean separation between atoms (from the ideal gas law).

At $T=300\ \text{K}$ and $P=1\ \text{atm}=1.013\times10^{5}\ \text{Pa}$ , the volume occupied per atom is

\frac{V}{N} = \frac{kT}{P} = \frac{(1.38\times10^{-23})(300)}{1.013\times10^{5}} \approx 4.09\times10^{-26}\ \text{m}^3

Taking the cube root gives the typical spacing between neighbouring atoms:

$d = \left(\frac{V}{N}\right)^{1/3} = (4.09\times10^{-26})^{1/3} \approx 3.45\times10^{-9}\ \text{m} = 34.5\ \text{Å}$ …

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Q.Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C27\ °\text{C}27 °C) and 1 atm1\ \text{atm}1 atm pressure; and compare it with the mean separation between two atoms under these conditions.

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Q.Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature ( $27\ °\text{C}$ ) and $1\ \text{atm}$ pressure; and compare it with the mean separation between two atoms under these conditions.

Unlock everything free for 14 days