Animated Solution for Physics - Dual Nature of Matter and Radiation: In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of $7.5\times 10^{-12}\text{ m}$, the minimum electron energy required is close to

Electron Microscopes and de Broglie Waves

Have you ever wondered why optical microscopes have a fundamental limit to how much they can magnify? It all comes down to the nature of light. In wave optics, diffraction limits our ability to distinguish two close objects. The Rayleigh criterion dictates that the minimum resolvable distance is roughly proportional to the wavelength of the wave used to observe them. Visible light has a wavelength ranging from $400$ to $700$ nanometers. If you want to see something smaller than that—like the atomic structure of a crystal—you need a wave with a much, much smaller wavelength.

This is where Louis de Broglie's brilliant hypothesis comes into play. He proposed that matter, just like light, exhibits wave-like properties. By accelerating electrons, we can give them a tremendous amount of momentum, which in turn gives them an incredibly tiny wavelength. This is the core principle behind the electron microscope!

In this problem, we are tasked with finding the minimum electron energy required to resolve a width of $d=7.5\times 10^{-12}\text{ m}$. According to the resolution criterion, the wavelength $\lambda$ of the electrons must be on the order of this distance:

Now, how do we connect this wavelength to the energy of the electron? We start with the de Broglie relation, which links wavelength to momentum $p$:

Next, we recall the classical relationship between kinetic energy $K$ and momentum for a non-relativistic particle:

Substituting our expression for momentum into the kinetic energy equation, we get our master formula:

Now comes the part where we must be incredibly careful with our scientific notation. Let's substitute the known constants: Planck's constant $h=6.6\times 10^{-34}\text{ J}\cdot \text{s}$, the mass of an electron $m=9.1\times 10^{-31}\text{ kg}$, and our target wavelength $\lambda =7.5\times 10^{-12}\text{ m}$.

Squaring the terms in the numerator and denominator:

Dividing the coefficients and subtracting the exponents:

In atomic physics, Joules are often too large and cumbersome to work with. We prefer electron-volts (eV). To convert our energy from Joules to eV, we divide by the elementary charge $e=1.6\times 10^{-19}\text{ C}$:

This can be neatly written in kilo-electron-volts (keV):

Looking at our options, the closest value is $25\text{ keV}$. The slight discrepancy arises from the exact values of the constants used (like using $6.626$ instead of $6.6$ for $h$), but in the context of a multiple-choice question, $25\text{ keV}$ is the clear and unambiguous answer.

By mastering this relationship between energy, momentum, and wavelength, you unlock the mathematical foundation of modern quantum microscopy!