Animated Solution for Physics - Dual Nature of Matter and Radiation: An electron of mass $m_e$ and a proton of mass $m_p$ are accelerated through the same potential difference. The ratio of the de-Broglie wavelength associated with the electron to that with the proton is

Have you ever wondered what happens when you take the two most fundamental building blocks of our universe—an electron and a proton—and put them in a race track powered by an electric field? This problem takes us on a beautiful journey into the quantum realm, where particles behave like waves.

Imagine you are standing in a lab. You have an electron and a proton, both initially at rest. You flip a switch, and both are accelerated through the exact same potential difference, $V$. Our goal is to find out how their quantum wavelengths—the de-Broglie wavelengths—compare at the end of this acceleration.

To solve this, we need to bridge the gap between the macroscopic world of voltages and the quantum world of matter waves. We start with the legendary de-Broglie equation:

Here, $h$ is Planck's constant and $p$ is the momentum of the particle. But we don't know their momenta directly. We do, however, know about their energy. The kinetic energy $K$ of a particle is related to its momentum by the equation $K=\frac{p^2}{2m}$, which means $p=\sqrt{2mK}$. Substituting this into our wave equation gives:

Now, how much kinetic energy do they gain? When a particle with charge $q$ is accelerated through a potential difference $V$, the electrical work done on it is $qV$. By the work-energy theorem, this becomes its kinetic energy ($K=qV$). Let's plug this into our equation:

This is our master equation. It beautifully connects the quantum wavelength $\lambda$ to the macroscopic accelerating potential $V$.

In physics, especially in competitive exams like JEE, the secret to speed and accuracy lies in identifying what doesn't change. Let's look at our master equation and compare the electron and the proton.

1. Potential Difference ($V$): The problem explicitly states they are accelerated through the same potential difference. 2. Charge ($q$): The electron has a charge of $-e$ and the proton has a charge of $+e$. Since we only care about the magnitude of energy gained, $q=e$ for both. 3. Constants: Planck's constant $h$ and the number $2$ are, of course, universal constants.

Since $h$, $2$, $q$, and $V$ are all identical for both particles, we can strip them away to reveal the core mathematical relationship:

This tells us a profound physical truth: for particles accelerated through the same potential, the heavier the particle, the smaller its quantum wavelength.

Now, finding the ratio is a breeze. We want the ratio of the electron's wavelength (${\lambda }_e$) to the proton's wavelength (${\lambda }_p$). Because of the inverse square root relationship, the masses will flip in the ratio:

And there we have it! Because the proton is roughly 1836 times more massive than the electron, the electron's wavelength will be significantly larger—about 43 times larger, to be precise.

Always remember, before you start plugging in numbers or doing complex algebra, look for the constants. Proportionality is your best friend in physics!