Two particles A and B of de Broglie wavelengths $\lambda_1$ and $\lambda_2$ combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional.)

Two particles A and B of de Broglie wavelengths λ₁ and λ₂ combine to form a particle C. The process conserves

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\lambda_1

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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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The de Broglie wavelength of the combined particle C is $\boxed{\frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}}$ when the two particles move in the same direction, and $\boxed{\frac{\lambda_1 \lambda_2}{|\lambda_1 - \lambda_2|}}$ when they move in opposite directions. The key is that momentum is conserved, and de Broglie wavelength is inversely proportional to momentum.

The Core Idea: Wavelength as Momentum

The de Broglie hypothesis tells us that every moving particle has a wavelength associated with it, given by:

\lambda = \frac{h}{p}

where $h$ is Planck's constant and $p$ is the linear momentum. This is the single most important relation here. Notice the inverse relationship: a larger momentum means a shorter wavelength, and vice versa.

Since the problem states that momentum is conserved during the combination, the total momentum before the event must equal the momentum of the resulting particle C. Therefore, if we can express the momenta of A and B in terms of their wavelengths, we can add them (vectorially, since this is one-dimensional motion) to find the momentum of C, and then convert that back into a wavelength.

Step-by-Step Solution

1. Express the momenta of A and B.

From the de Broglie relation, the momentum of a particle is $p = \frac{h}{\lambda}$ . So for particles A and B:

p_1 = \frac{h}{\lambda_1}, \quad p_2 = \frac{h}{\lambda_2}

2. Account for direction in one dimension.

The problem says the motion is one-dimensional. This means the particles are moving along the same line, but they could be moving in the same direction or in opposite directions. Momentum is a vector, so we must assign a sign. Let’s take the positive direction as the direction of motion of particle A. Then:

If B moves in the same direction as A, its momentum is $+p_2$ .
If B moves in the opposite direction to A, its momentum is $-p_2$ .

3. Apply conservation of momentum.

The total momentum before combination equals the momentum of C after combination:

p_C = p_1 + p_2

But careful: this is a vector sum. In one dimension, we just add with appropriate signs.

Watch out

A common mistake is to treat wavelengths as if they add directly. They don't. Wavelengths are inversely proportional to momentum, so you must first convert to momentum, add, then convert back.

4. Find the wavelength of C.

The de Broglie wavelength of C is $\lambda_C = \frac{h}{p_C}$ . So:

\lambda_C = \frac{h}{p_1 + p_2} = \frac{h}{\frac{h}{\lambda_1} + \frac{h}{\lambda_2}}

Factor out $h$ from the denominator:

\lambda_C = \frac{h}{h\left(\frac{1}{\lambda_1} + \frac{1}{\lambda_2}\right)} = \frac{1}{\frac{1}{\lambda_1} + \frac{1}{\lambda_2}}

Now simplify the fraction:

$\frac{1}{\lambda_C} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$ …

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Q.Two particles A and B of de Broglie wavelengths $\lambda_1$ and $\lambda_2$ combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional.)

Concept understanding — De Broglie Wavelength

De Broglie Wavelength: When Particles Start Acting Like Waves

The Intuition

The Precise Statement

What This Means Physically

Worked Example

Why this formula?

De Broglie Wavelength: Why Matter Has a Wave Nature

The Core Insight: Symmetry in Nature

Why This Formula Makes Sense: A Simple Derivation

The Core Idea: Wavelength as Momentum

Step-by-Step Solution

Method: Converting Between Wavelength and Momentum for Combination/Collision Problems

Steps

Q.Two particles A and B of de Broglie wavelengths λ1\lambda_1λ1​ and λ2\lambda_2λ2​ combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional.)

The Core Idea: Wavelength as Momentum

Step-by-Step Solution

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Q.Two particles A and B of de Broglie wavelengths $\lambda_1$ and $\lambda_2$ combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional.)

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