Animated Solution for Physics - Dual Nature of Matter and Radiation: An electron of mass $m$ and a photon have same energy $E$. The ratio of wavelength of electron to that of photon is ($c$ being the velocity of light)

Matter vs. Light

Imagine a cosmic race between two fundamentally different entities. On one side, we have an electron—a tiny, tangible particle with a definite mass $m$. On the other side, we have a photon—a massless, ethereal packet of pure light.

Despite their differences, they share one crucial characteristic in this scenario: they both possess the exact same energy $E$.

Our mission is to find the ratio of their wavelengths, $\frac{{\lambda }_e}{{\lambda }_p}$. To do this, we need to bridge the gap between classical mechanics and quantum physics.

Let's first focus on the electron. According to Louis de Broglie's revolutionary hypothesis, every moving particle has an associated wave.

The wavelength of this matter wave is given by Planck's constant divided by the particle's momentum: ${\lambda }_e=\frac{h}{p}$

But we are given the electron's kinetic energy $E$, not its momentum. We know the classic relationship between kinetic energy and momentum is $E=\frac{p^2}{2m}$.

Rearranging this for momentum gives us $p=\sqrt{2mE}$. Substituting this back into the de Broglie equation, we get the electron's wavelength: ${\lambda }_e=\frac{h}{\sqrt{2mE}}$

Now, let's turn our attention to the photon. For a quantum of light, the relationship between energy and wavelength is much more direct.

The energy of a photon is given by the famous Planck-Einstein relation: $E=\frac{hc}{{\lambda }_p}$

Where $c$ is the speed of light. By simply rearranging this formula, we can isolate the photon's wavelength: ${\lambda }_p=\frac{hc}{E}$

Notice how the fundamental nature of these two entities leads to very different wavelength formulas!

Finding the Ratio

We have our two wavelengths. Now, it's time to find their ratio by dividing the electron's wavelength by the photon's wavelength.

This looks like a messy fraction, but let's clean it up. When we multiply the numerator by the reciprocal of the denominator, something beautiful happens.

The Planck's constant $h$ cancels out completely! This cancellation is a profound reminder that both matter and light are governed by the same underlying quantum rules.

We are left with: $\frac{{\lambda }_e}{{\lambda }_p}=\frac{E}{c\sqrt{2mE}}$

We are almost at the finish line. We just need to simplify the expression to match the given options.

Notice that we have an $E$ in the numerator and a $\sqrt{E}$ in the denominator. Since $E$ divided by $\sqrt{E}$ is simply $\sqrt{E}$, we can bring the $E$ inside the square root.

Simplifying the fraction inside the root gives us: $\frac{{\lambda }_e}{{\lambda }_p}=\frac{1}{c}\sqrt{\frac{E}{2m}}$

Which can also be written using a fractional exponent: $\frac{{\lambda }_e}{{\lambda }_p}=\frac{1}{c}{\left(\frac{E}{2m}\right)}^{1/2}$

As a final thought, notice that the ratio is proportional to $\sqrt{E}$. This means that as the energy of both particles increases, the ratio increases. The photon's wavelength shrinks much faster than the electron's wavelength at higher energies!