Animated Solution for Physics - Dual Nature of Matter and Radiation: An electron (of mass $m$) and a photon have the same energy $E$ in the range of a few electron volt. The ratio of the de Broglie wavelength associated with the electron and the wavelength of the photon is ($c=$ speed of light in vacuum)

The Tale of Two Wavelengths: Electron vs. Photon

Have you ever wondered how a massive particle like an electron compares to a massless packet of light like a photon when they both possess the exact same energy? This classic JEE problem invites us to explore the dual nature of matter and radiation by calculating the ratio of their wavelengths. Let's dive into the quantum world and unravel this elegant relationship.

Let's start with the electron. It is a fundamental particle with a rest mass $m$. The problem states that it has a kinetic energy $E$. To find its wavelength, we need to use the de Broglie hypothesis, which connects the momentum of a particle to its wavelength.

First, we relate the kinetic energy to momentum. For a non-relativistic particle, the kinetic energy is given by:

Rearranging this to solve for momentum $p$, we get:

Now, according to de Broglie, the wavelength ${\lambda }_e$ associated with this electron is Planck's constant $h$ divided by its momentum:

Next, let's look at the photon. A photon is a quantum of electromagnetic radiation, and it travels at the speed of light $c$. Its energy is determined by the famous Planck-Einstein relation:

Where ${\lambda }_p$ is the wavelength of the photon. We can easily rearrange this equation to express the wavelength in terms of energy:

Notice the stark difference here: the electron's wavelength depends on the square root of its energy, while the photon's wavelength is inversely proportional to its energy directly.

Now that we have the expressions for both wavelengths, the final step is to find their ratio. We simply divide ${\lambda }_e$ by ${\lambda }_p$:

When we multiply by the reciprocal of the denominator, a beautiful thing happens—Planck's constant $h$ cancels out completely!

To simplify this further, we can write $E$ in the numerator as $\sqrt{E^2}$. Bringing everything under a single square root gives:

This matches option (d). The ratio is $\frac{1}{c}{\left(\frac{E}{2m}\right)}^{1/2}$.

Before we conclude, let's address a subtle but crucial detail in the problem statement: "in the range of a few electron volts". Why did the examiner include this phrase?

This is a safety net. The rest mass energy of an electron is about $0.511\text{ MeV}$ (or $511,000\text{ eV}$). Because our energy $E$ is only a few $\text{eV}$, it is vastly smaller than the rest mass energy ($E\ll m{c}^2$). This guarantees that the electron is moving at a non-relativistic speed, making our classical kinetic energy formula $E=\frac{p^2}{2m}$ perfectly valid.

If the energy were in the $\text{MeV}$ range, the electron would be moving close to the speed of light. We would then have to use the relativistic energy-momentum relation $E_{total}^2=(pc{)}^2+(m{c}^2{)}^2$, and our entire derivation would change! Always keep an eye out for these subtle constraints in JEE physics problems; they separate the good students from the great ones.