Animated Solution for Physics - Dual Nature of Matter and Radiation: Two particles move at right angle to each other. Their de-Broglie wavelengths are ${\lambda }_1$ and ${\lambda }_2$, respectively. The particles suffer perfectly inelastic collision. The de-Broglie wavelength $\lambda$ of the final particle, is given by

Imagine two particles hurtling through space, completely unaware of each other, moving along paths that intersect at a perfect right angle. This is the classic setup of our problem. We are told that these particles have de-Broglie wavelengths of ${\lambda }_1$ and ${\lambda }_2$.

Before we even think about the collision, we need to translate these wavelengths into a dynamic property that we can work with. This is where the genius of Louis de Broglie comes in. His famous relation, $\lambda =\frac{h}{p}$, bridges the gap between the wave-like nature of a particle (its wavelength $\lambda$) and its particle-like nature (its momentum $p$).

By rearranging this formula, we can express the initial momentum of each particle. For the first particle, its momentum is $p_1=\frac{h}{{\lambda }_1}$. Similarly, for the second particle, its momentum is $p_2=\frac{h}{{\lambda }_2}$. Now we have our two momentum vectors, ${\vec{p}}_1$ and ${\vec{p}}_2$, ready for the impending collision.

The problem states that the particles undergo a perfectly inelastic collision. In the physical world, this means that upon impact, the two particles do not bounce off each other. Instead, they stick together, merging into a single, combined mass that moves off in a new direction.

While a significant amount of kinetic energy is lost in such a violent collision (transformed into heat or internal energy), there is one fundamental law of the universe that remains unbroken: the conservation of linear momentum. The total momentum of the system before the collision must exactly equal the total momentum after the collision.

Mathematically, this means the final momentum vector, ${\vec{p}}_{\text{net}}$, is simply the vector sum of the initial momenta: ${\vec{p}}_{\text{net}}={\vec{p}}_1+{\vec{p}}_2$.

Because our initial particles were moving at right angles to each other, their momentum vectors are perpendicular. This makes finding the magnitude of the final momentum incredibly straightforward. We can just use the Pythagorean theorem! The magnitude of the net momentum is given by $p_{\text{net}}=\sqrt{p_1^2+p_2^2}$.

Now, let's bring it all together. We substitute our earlier expressions for $p_1$ and $p_2$ into our Pythagorean equation. This gives us $p_{\text{net}}=\sqrt{{\left(\frac{h}{{\lambda }_1}\right)}^2+{\left(\frac{h}{{\lambda }_2}\right)}^2}$.

But we aren't looking for the final momentum; we want the final de-Broglie wavelength, let's call it $\lambda$. Applying the de-Broglie relation one last time to our combined particle, we know that $p_{\text{net}}=\frac{h}{\lambda }$.

Equating these two expressions for $p_{\text{net}}$, we get:

To clean this up and reveal the final elegant relationship, we square both sides of the equation to eliminate the square root. This yields:

Notice how the $h^2$ term is present in every single part of the equation? We can beautifully divide the entire equation by $h^2$, causing it to cancel out completely. We are left with our final, pristine answer:

This result elegantly shows how the wave properties of the particles combine under the strict rules of momentum conservation.