(a) Obtain the de Broglie wavelength of a neutron of kinetic energy $150\ \text{eV}$. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. ($m_n = 1.675 \times 10^{-27}\ \text{kg}$) (b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature ($27\ °\text{C}$). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150eV. As you have seen in Exercise 11.31,

Additional Exercises · 11.32

150\ \text{eV}

(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (

27\ °\text{C}

). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.

CBSENCERTSubjective· 3mImportance★★★★★

42% · 35/83 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 $\lambda = h/p$ . …

Start your 14-day free trial to unlock the full solution →

A 150 eV neutron's wavelength (≈0.023 Å) is far shorter than inter-atomic spacing, so it's useless for diffraction — unlike a 150 eV electron. Only after being thermalised (slowed to room-temperature thermal energies) does a neutron's wavelength grow to ≈1.45 Å, comparable to lattice spacing, making it suitable.

(a) de Broglie wavelength of a 150 eV neutron.

KE = 150\ \text{eV} = 150\times1.6\times10^{-19} = 2.4\times10^{-17}\ \text{J}

p = \sqrt{2m_nKE} = \sqrt{2(1.675\times10^{-27})(2.4\times10^{-17})} = \sqrt{8.04\times10^{-44}} \approx 2.836\times10^{-22}\ \text{kg m/s}

\lambda = \frac{h}{p} = \frac{6.63\times10^{-34}}{2.836\times10^{-22}} \approx 2.34\times10^{-12}\ \text{m} = 0.0234\ \text{Å}

Since useful crystal diffraction needs $\lambda$ of order the inter-atomic spacing (~1 Å, as noted in Exercise 11.31), and this neutron's wavelength (0.0234 Å) is about 40 times too short, a 150 eV neutron beam would not be suitable for diffraction — because a neutron is nearly 1840 times heavier than an electron of the same energy, it acquires a far larger momentum, and hence a far shorter de Broglie wavelength, for the same kinetic energy.

(b) de Broglie wavelength of a thermal neutron at $T=300\ \text{K}$ .

KE = \frac{3}{2}kT = 1.5(1.38\times10^{-23})(300) = 6.21\times10^{-21}\ \text{J}

p = \sqrt{2m_nKE} = \sqrt{2(1.675\times10^{-27})(6.21\times10^{-21})} = \sqrt{2.080\times10^{-47}} \approx 4.56\times10^{-24}\ \text{kg m/s}

$\lambda = \frac{h}{p} = \frac{6.63\times10^{-34}}{4.56\times10^{-24}} \approx 1.45\times10^{-10}\ \text{m} = 1.45\ \text{Å}$ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set ANNUAL1 mark
$Q. What is the ratio of de-Broglie wavelength associated with two electron beams A and B accelerated through 25 V and 36 V respectively ?$
›Reveal solutionSolution

de Broglie wavelength for an electron accelerated through potential V: $\lambda \propto 1/\sqrt V$ .

For an electron accelerated through a potential difference V, $\lambda = \dfrac{h}{\sqrt{2meV}}$ , so $\lambda \propto \dfrac{1}{\sqrt V}$ .

…

Start your 14-day free trial to see the full solution →

CBSE 2019Set 55/5/11 mark

Q. Plot a graph of the de-Broglie wavelength associated with a proton versus its momentum.

›Reveal solution

The de-Broglie wavelength $\lambda$ of a proton is inversely proportional to its momentum $p$ , following $\lambda = h/p$ . The graph is a rectangular hyperbola in the first quadrant — a smooth curve that falls steeply at low momentum and flattens as momentum increases.

The idea is beautifully simple. Louis de Broglie proposed that every moving particle has a wavelength associated with it, given by $\lambda = h/p$ , where $h$ is Planck’s constant and $p$ is the linear momentum. For a proton, this relation holds exactly — no approximations, no mass dependence once you write it in terms of $p$ .

Why does this matter? Because the graph tells you something immediate: as you speed up a proton (increase its momentum), its wavelength shrinks. At low speeds, the wavelength is large; at very high speeds, it becomes tiny. The shape is a rectangular hyperbola — the same curve you get for $xy = \text{constant}$ .

Let’s build the graph step by step.

Write the relation. The de-Broglie wavelength is

\lambda = \frac{h}{p}

where $h = 6.63 \times 10^{-34} \ \text{J·s}$ . For a proton, $p = mv$ (non-relativistic) or $p = \gamma mv$ (relativistic), but the $\lambda$ vs $p$ relation itself is universal — it doesn’t care about mass or speed.

Identify the axes.
- Horizontal axis: momentum $p$ (in kg·m/s, say).
- Vertical axis: wavelength $\lambda$ (in metres). Both are positive quantities, so the graph lies entirely in the first quadrant.
Understand the shape.

The equation $\lambda = h/p$ is of the form $y = c/x$ , a rectangular hyperbola. Key features:
- As $p \to 0^+$ , $\lambda \to \infty$ — the curve shoots up near the vertical axis.
- As $p \to \infty$ , $\lambda \to 0$ — the curve approaches the horizontal axis asymptotically.
- The product $\lambda p = h$ is constant, so every point on the curve satisfies this.

Plot a few points to guide the sketch.

Let’s pick some convenient momentum values and compute $\lambda$ :

$p$ (kg·m/s)	$\lambda = h/p$ (m)
$1 \times 10^{-24}$	$6.63 \times 10^{-10}$
$2 \times 10^{-24}$	$3.32 \times 10^{-10}$
$5 \times 10^{-24}$	$1.33 \times 10^{-10}$
$1 \times 10^{-23}$	$6.63 \times 10^{-11}$

Notice: doubling $p$ halves $\lambda$ — that’s the inverse proportionality in action. …

Start your 14-day free trial to see the full solution →

←Previous question11.31 — Crystal diffraction experiments can be performed using X-rays, or elec…→Next question11.33 — An electron microscope uses electrons accelerated by a voltage of 50kV…

🎓Unlock everything free for 14 days

Unlock everything free for 14 days