Animated Solution for Physics - Dual Nature of Matter and Radiation: The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of $100\text{ V}$. What should nearly be the ratio of their wavelengths? ($m_p=1.00727\text{ u}$, $m_e=0.00055\text{ u}$)

Electron vs. Proton

Imagine an electron and a proton, both standing at the starting line of a particle accelerator. We flip the switch, applying a potential difference of $100\text{ V}$. The electron darts towards the positive plate, while the proton lumbers towards the negative plate. As they accelerate, they gain kinetic energy, and according to quantum mechanics, they begin to exhibit wave-like properties. But whose wave is longer? Let's dive into the mathematics of matter waves to find out.

To find the ratio of their de-Broglie wavelengths, we need a formula that connects the wavelength $\lambda$ to the accelerating potential $V$. We start with the fundamental de-Broglie relation:

We know that the momentum $p$ is related to kinetic energy $K$ by $p=\sqrt{2mK}$. Furthermore, the kinetic energy gained by a particle of charge $q$ accelerated through a potential $V$ is $K=qV$. Substituting this into our momentum equation gives us the master equation for this problem:

Now, let's look closely at this formula and identify the constants for our specific scenario. Planck's constant $h$ is a universal constant. The accelerating potential $V$ is given as $100\text{ V}$ for both particles. Finally, the magnitude of the charge $q$ is identical for both an electron and a proton ($q_e=q_p=e$).

Since $h$, $q$, and $V$ are all constant, we can establish a beautiful proportionality. The de-Broglie wavelength is inversely proportional to the square root of the particle's mass:

This tells us intuitively that the heavier the particle, the shorter its wavelength. Because we want the ratio of the electron's wavelength to the proton's wavelength, we can set up the following relation:

Now comes the execution phase. We substitute the given mass values into our ratio equation. The mass of the proton is $m_p=1.00727\text{ u}$, and the mass of the electron is $m_e=0.00055\text{ u}$. Notice that we don't need to convert atomic mass units (u) to kilograms because the units will perfectly cancel out in the ratio.

To make the division easier, we can shift the decimal point five places to the right for both numbers:

Finally, we need to find the square root of $1831.4$. We can estimate this by recalling perfect squares: $40^2=1600$ and $45^2=2025$. Our number is roughly in the middle, slightly closer to $1600$. If we test $43^2$, we get $1849$, which is incredibly close to $1831.4$.

Rounding to the nearest integer, we get $43$. Therefore, the ratio of their wavelengths is $43:1$.

A Fascinating Takeaway: Did you notice that the $100\text{ V}$ potential completely vanished from our calculation? This means whether you accelerate these particles with $100\text{ V}$, $1000\text{ V}$, or a million volts, the ratio of their de-Broglie wavelengths will always remain $43:1$. It is an intrinsic property dictated solely by the ratio of their masses!