Animated Solution for Physics - Dual Nature of Matter and Radiation: A particle A of mass $m$ and charge $q$ is accelerated by a potential difference of $50\text{ V}$. Another particle B of mass $4m$ and charge $q$ is accelerated by a potential difference of $2500\text{ V}$. The ratio of de-Broglie wavelengths $\frac{{\lambda }_A}{{\lambda }_B}$ is close to

Welcome to the fascinating world of quantum mechanics, where particles behave like waves! In 1924, Louis de Broglie proposed a revolutionary idea: just as light waves can exhibit particle-like properties (photons), particles of matter can exhibit wave-like properties. This dual nature is beautifully captured by the de-Broglie wavelength formula.

In this problem, we are going to explore how the de-Broglie wavelength of a charged particle changes when it is accelerated by an electric potential. Let's dive in!

Imagine we have two distinct particles, let's call them Particle A and Particle B.

Particle A is relatively light, with a mass of $m$, and carries a charge $q$. It is placed in an electric field and accelerated by a potential difference of $V_A=50\text{ V}$.

Particle B, on the other hand, is much heavier. Its mass is $4m$, which is four times that of Particle A. It carries the exact same charge $q$, but it is subjected to a much stronger accelerating potential difference of $V_B=2500\text{ V}$.

Our goal is to find the ratio of their resulting de-Broglie wavelengths, $\frac{{\lambda }_A}{{\lambda }_B}$.

The fundamental equation for the de-Broglie wavelength is:

where $h$ is Planck's constant and $p$ is the momentum of the particle. However, we aren't given the momentum directly. Instead, we know the accelerating potential.

When a particle of charge $q$ is accelerated from rest through a potential difference $V$, the electrical work done on it is converted entirely into kinetic energy ($K$). Therefore, $K=qV$.

We also know the relationship between kinetic energy and momentum: $K=\frac{p^2}{2m}$. Rearranging this for momentum gives us $p=\sqrt{2mK}$.

Substituting $K=qV$ into our momentum equation, we get $p=\sqrt{2mqV}$. Finally, plugging this back into the de-Broglie wavelength formula yields our master equation:

Now, let's set up the ratio $\frac{{\lambda }_A}{{\lambda }_B}$.

When we divide these two fractions, the constants $h$ and $2$ beautifully cancel out. Because the terms are in the denominator, the ratio flips, giving us a clean inverse relationship inside a single square root:

It's time to substitute the specific values given in the problem: - $m_B=4m$ and $m_A=m$ - $q_B=q$ and $q_A=q$ - $V_B=2500\text{ V}$ and $V_A=50\text{ V}$

Plugging these in:

The masses ($m$) and charges ($q$) cancel out perfectly. We are left with:

Let's break down the square root to make the calculation easier:

We know that $\sqrt{4}=2$. For $\sqrt{50}$, we can write it as $\sqrt{25\cdot 2}$, which simplifies to $5\sqrt{2}$.

Finally, using the standard approximation $\sqrt{2}\approx 1.414$:

The ratio of their de-Broglie wavelengths is 14.14. This perfectly matches option (d).

Problems like this might look intimidating at first glance, but by carefully setting up the ratios, the complex variables often cancel out, leaving you with a simple and elegant calculation!