Animated Solution for Physics - Dual Nature of Matter and Radiation: An electron moving with speed $v$ and a photon moving with speed $c$, have same de-Broglie wavelength. The ratio of kinetic energy of electron to that of photon is

Imagine a fascinating cosmic race: an electron zooming through space with a velocity $v$, and a photon blazing past it at the ultimate speed limit of the universe, $c$. The problem presents us with a beautiful quantum mechanical constraint—despite their vastly different natures, both particles possess the exact same de-Broglie wavelength, ${\lambda }_e={\lambda }_p=\lambda$.

Our mission is to find the ratio of the kinetic energy of the electron to the energy of the photon. At first glance, this might seem like a heavy algebraic exercise involving masses, velocities, and Planck's constant. However, by looking through the lens of momentum, the solution reveals itself with elegant simplicity.

The key to unlocking this problem lies in the fundamental de-Broglie relation, which bridges the wave and particle natures of matter:

Since the problem explicitly states that their wavelengths are equal (${\lambda }_e={\lambda }_p$), it mathematically guarantees that their momenta must also be perfectly equal. Let's denote this shared, common momentum as $p$:

This simple equality is the secret weapon that will slice through the complex algebra later on.

Let's analyze the electron first. The classical formula for the kinetic energy of a particle with mass $m_e$ and velocity $v$ is:

While we could use this directly, it's strategically much smarter to express this energy in terms of momentum. We can rewrite the $v^2$ term by splitting it:

Recognizing that mass times velocity ($m_{e}v$) is exactly the momentum of the electron ($p_e$), we arrive at a much more useful form:

Now, let's turn our attention to the photon. The energy of a photon is given by the Planck-Einstein relation:

Notice the $\frac{h}{\lambda }$ term embedded within this equation? From the de-Broglie relation we discussed earlier, $\frac{h}{\lambda }$ is precisely the momentum of the photon ($p_p$). Substituting this in, we get a beautifully simple expression for the photon's energy:

We are now perfectly equipped to find the requested ratio. Let's set up the fraction by dividing the electron's kinetic energy by the photon's energy:

Here is where the magic happens. Remember our initial deduction that $p_e=p_p=p$? Because the momenta are identical, the $p$ terms in the numerator and the denominator completely cancel each other out!

And just like that, the complex physics collapses into a clean, elegant fraction. The ratio of their kinetic energies is simply $\frac{v}{2c}$. This problem is a stellar example of how choosing the right physical variables—in this case, momentum—can turn a potentially messy calculation into a smooth and satisfying derivation.