Animated Solution for Physics - Dual Nature of Matter and Radiation: The potential energy of a particle varies as $U(x)=E_0$ for $0\le x\le 1$ $=0$ for $x>1$ For $0\le x\le 1$, de-Broglie wavelength is ${\lambda }_1$ and for $x>1$ the de-Broglie wavelength is ${\lambda }_2$. Total energy of the particle is $2E_0$. Find $\frac{{\lambda }_1}{{\lambda }_2}$.

Navigating Energy Landscapes

Imagine a skateboarder rolling along a track. The track has a raised platform in one section and is flat everywhere else. If the skateboarder has enough energy to get onto the platform, they will move slower while on it, and speed up when they roll off. This is exactly what is happening to our quantum particle in this problem!

We are given a potential energy landscape $U(x)$ that acts like our raised platform. Between $x=0$ and $x=1$, the potential energy is $E_0$. Everywhere past $x=1$, the potential energy drops to zero.

The total energy of our particle is a constant $E=2E_0$.

In classical mechanics, the total energy is the sum of kinetic and potential energy:

Because the total energy $E$ is constant, whenever the potential energy $U$ changes, the kinetic energy $K$ must adjust to compensate.

Let's look at the first region ($0\le x\le 1$). Here, the potential energy is $U_1=E_0$. We can find the kinetic energy $K_1$ by subtracting the potential energy from the total energy:

Now, let's look at the second region ($x>1$). The potential energy drops to $U_2=0$. Again, we find the kinetic energy $K_2$:

Notice how the particle speeds up in the second region because it has converted its potential energy into kinetic energy!

In the quantum world, moving particles behave like waves. The wavelength of this matter wave is given by the de-Broglie relation:

We can express the momentum $p$ in terms of kinetic energy $K$ using the relation $p=\sqrt{2mK}$. Substituting this into our de-Broglie equation gives:

This tells us a crucial fact: the de-Broglie wavelength is inversely proportional to the square root of the kinetic energy.

We need to find the ratio of the wavelengths in the two regions, $\frac{{\lambda }_1}{{\lambda }_2}$. Using our inverse proportionality rule, we can write:

Now, we simply substitute the kinetic energies we calculated earlier:

The $E_0$ terms cancel out beautifully, leaving us with our final answer:

This means the wavelength in the first region is $\sqrt{2}$ times longer than in the second region, perfectly reflecting the fact that the particle is moving slower on the "raised platform"!