Animated Solution for Physics - Dual Nature of Matter and Radiation: A beam of electrons of energy $E$ scatters from a target having atomic spacing of $1\text{A˚}$. The first maximum intensity occurs at $\theta =60^{\circ }$. Then, $E$ (in eV) is ......... . (Given, Planck's constant, $h=6.64\times 10^{-34}\text{ Js}$, $1\text{ eV}=1.6\times 10^{-19}\text{ J}$ and electron mass, $m=9.1\times 10^{-31}\text{ kg}$.)

## Unveiling the Energy of Electrons: A Journey through Bragg's Law and Matter Waves

Have you ever wondered how we can "see" the invisible structure of a crystal? The answer lies in one of the most profound discoveries of the 20th century: the wave nature of matter. In this problem, we are not dealing with light or X-rays, but with a beam of electrons. Yet, these electrons are behaving exactly like waves, diffracting off the atomic planes of a crystal target.

Let's embark on a thrilling journey to uncover the kinetic energy of these electrons by beautifully merging two monumental concepts in physics: Bragg's Law of diffraction and the de Broglie wavelength hypothesis.

Imagine a highly ordered crystal lattice. The atoms in this crystal are arranged in perfectly parallel planes, separated by a regular atomic spacing, $d=1\text{ A˚}$ (which is $10^{-10}\text{ m}$).

Now, visualize a beam of electrons firing at this target. When these electrons hit the crystal, they don't just bounce off randomly like billiard balls. Because of their quantum mechanical wave nature, they scatter and interfere with each other. The problem states that the first maximum intensity occurs at a glancing angle of $\theta =60^{\circ }$. This "maximum intensity" is the hallmark of constructive interference, where the scattered waves perfectly align crest-to-crest.

To find the condition for this constructive interference, we invoke Bragg's Law. For the first-order maximum ($n=1$), the path difference between waves scattering off adjacent planes must equal exactly one wavelength:

But wait, what is the wavelength $\lambda$ of an electron? This is where Louis de Broglie steps in. He proposed that any moving particle has an associated wavelength, given by Planck's constant $h$ divided by its momentum $p$. Since kinetic energy $E=\frac{p^2}{2m}$, we can express the momentum as $p=\sqrt{2mE}$. Therefore, the de Broglie wavelength is:

Now, we have two different perspectives on the same wavelength $\lambda$. Let's bridge them together by equating the two expressions:

Our goal is to find the kinetic energy $E$. To liberate $E$ from the square root, we must square both sides of the equation. This is a crucial algebraic step:

Rearranging this to make $E$ the subject, we arrive at our master equation:

With our master equation ready, it's time to plug in the numbers. This is where many students make silly mistakes, so we must be extremely disciplined with our units. Everything must be in standard SI units (meters, kilograms, seconds, Joules).

Let's list our given values: - Planck's constant, $h=6.64\times 10^{-34}\text{ Js}$ - Mass of an electron, $m=9.1\times 10^{-31}\text{ kg}$ - Atomic spacing, $d=10^{-10}\text{ m}$ - Glancing angle, $\theta =60^{\circ }$

Substituting these into our master equation:

Let's carefully evaluate the numerator and the denominator. The square of $6.64$ is approximately $44.0896$, and squaring the power of ten gives $10^{-68}$. For the denominator, remember that $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$, so ${\sin }^{2}60^{\circ }=\frac{3}{4}$.

We have successfully calculated the kinetic energy, but it is currently in Joules. The question specifically asks for the energy in electron-volts (eV).

To convert from Joules to eV, we must divide our result by the elementary charge, $1.6\times 10^{-19}\text{ C}$:

Handling the powers of ten, $10^{-17}/10^{-19}=10^2=100$. So the calculation simplifies to:

Rounding off to the nearest integer, we get our final, elegant answer:

Take a moment to appreciate what we just did. We used the macroscopic geometry of a crystal lattice to deduce the microscopic, quantum mechanical energy of a single electron. This seamless blend of classical wave optics and modern quantum physics is what makes this problem truly beautiful!