A particle P is formed due to a completely inelastic collision of particles x and y having de-Brogli

Q: Why do we subtract the momenta instead of adding them?

Momentum is a vector quantity. Because the problem explicitly states that the particles are moving in opposite directions, their momentum vectors have opposite signs. When we sum them up to find the net momentum, it mathematically becomes a subtraction of their magnitudes.

Q: Why is there an absolute value in the final answer?

We don't know whether particle $x$ or particle $y$ had the larger initial momentum. The absolute value ensures that the magnitude of the final momentum $p_P$ is positive, which is required because a wavelength $\lambda$ must be a positive physical distance.

Q: Does kinetic energy conserve in this collision?

No. The problem states it is a 'completely inelastic collision'. In such collisions, the particles stick together, and a maximum amount of kinetic energy is lost (converted into internal energy, heat, or deformation). Only linear momentum is conserved.

Q: Can we just add or subtract the wavelengths directly?

Absolutely not! Wavelengths do not add linearly like vectors. It is the momentum (which is inversely proportional to wavelength) that follows the rules of vector addition. Always convert wavelengths to momenta before applying conservation laws.

Q: What happened to the masses of the particles?

The beauty of the de-Broglie relation $\lambda = \frac{h}{p}$ is that it wraps mass and velocity together into a single variable: momentum. Because we are given the wavelengths, we already have all the information we need about their momenta, making their individual masses irrelevant to solving the problem.

Question

A particle $P$ is formed due to a completely inelastic collision of particles $x$ and $y$ having de-Broglie wavelengths $\lambda_x$ and $\lambda_y$, respectively. If $x$ and $y$ were moving in opposite directions, then the de-Broglie wavelength of $P$ is

Options:

(A) $\lambda_x - \lambda_y$

(B) $\frac{\lambda_x \lambda_y}{|\lambda_x - \lambda_y|}$

(C) $\frac{\lambda_x \lambda_y}{\lambda_x + \lambda_y}$

(D) $\lambda_x + \lambda_y$

Sigma Percentile · Accepted Answer

### The Setup: A Quantum Collision

Imagine two particles, $x$ and $y$, hurtling towards each other in the vast emptiness of space. They are on a direct collision course, moving in perfectly opposite directions. We don't know their masses or their velocities directly, but we are given something far more profound: their de-Broglie wavelengths, $\lambda_x$ and $\lambda_y$.

These particles are about to undergo a **completely inelastic collision**. This means that upon impact, they won't bounce off each other. Instead, they will fuse together, merging their identities to form a brand new, single entity which we will call particle $P$. Our mission is to uncover the de-Broglie wavelength of this newly born particle.

### The Bridge: Momentum and Wavelength

To bridge the gap between the initial state and the final state, we need a physical quantity that connects the quantum world of wavelengths to the classical world of collisions. That bridge is **momentum**.

The de-Broglie relation elegantly ties a particle's wavelength $\lambda$ to its momentum $p$ through Planck's constant $h$:

$$ \lambda = \frac{h}{p} $$

By rearranging this fundamental equation, we can express the momentum of any particle if we know its wavelength. Therefore, the initial momenta of our two particles are:

$$ p_x = \frac{h}{\lambda_x} \quad 	ext{and} \quad p_y = \frac{h}{\lambda_y} $$

### The Climax: Conservation of Momentum

Now comes the collision. The golden rule of physics that governs all collisions, whether elastic or inelastic, is the **Conservation of Linear Momentum**. The total momentum of the system before the collision must exactly equal the total momentum of the system after the collision.

Since particles $x$ and $y$ were moving in opposite directions, their momentum vectors point in opposite ways. When they fuse into particle $P$, the magnitude of the final momentum $p_P$ will simply be the absolute difference of their initial momentum magnitudes. We use the absolute value because we don't know which particle initially had more momentum, and the magnitude of momentum must be positive:

$$ p_P = |p_x - p_y| $$

Substituting the expressions we derived from the de-Broglie relation, we get the raw setup for our final particle's momentum:

$$ p_P = \left| \frac{h}{\lambda_x} - \frac{h}{\lambda_y} ight| $$

### The Resolution: Finding the New Wavelength

With the momentum of particle $P$ firmly in our grasp, finding its de-Broglie wavelength $\lambda_P$ is just one step away. We apply the de-Broglie relation one last time to the new particle:

$$ \lambda_P = \frac{h}{p_P} $$

Plugging in our expression for $p_P$, we get a slightly messy fraction:

$$ \lambda_P = \frac{h}{\left| \frac{h}{\lambda_x} - \frac{h}{\lambda_y} ight|} $$

Don't let the algebra intimidate you. We can easily factor out Planck's constant $h$ from the terms in the denominator. Since $h$ is a positive constant, it can be pulled right out of the absolute value signs:

$$ \lambda_P = \frac{h}{h \left| \frac{1}{\lambda_x} - \frac{1}{\lambda_y} ight|} $$

The $h$ in the numerator and the $h$ in the denominator cancel each other out beautifully. Now, we just find a common denominator for the fractions inside the absolute value:

$$ \lambda_P = \frac{1}{\left| \frac{\lambda_y - \lambda_x}{\lambda_x \lambda_y} ight|} $$

Flipping the fraction gives us our elegant final answer. Notice that $|\lambda_y - \lambda_x|$ is exactly the same as $|\lambda_x - \lambda_y|$:

$$ \lambda_P = \frac{\lambda_x \lambda_y}{|\lambda_x - \lambda_y|} $$

And there we have it! The wavelength of the fused particle is a harmonic-like combination of the original wavelengths.

Select Answer:

Visualized Solution

Initial Setup

de-Broglie Wavelength $λ$

Initial Momenta $p_{x}$ and $p_{y}$

The Collision

Conservation of Momentum

Final Momentum $p_{P}$

Wavelength of Particle $P$

Simplification

What if?

The Sigma Insight: Matter Waves and de Broglie Relation

The Setup

The Bridge

The Climax

The Resolution

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