For crystal diffraction experiments (discussed in Section...

For crystal diffraction experiments (discussed in Section $39.2 ),$ wavelengths on the order of 0.20 $\mathrm{nm}$ are often appropriate. Find the energy in electron volts for a particle with this wavclength if the particle is $(a)$ a photon; $(b)$ an clectron; $(c)$ an alpha particle $\left(m=6.64 \times 10^{-27} \mathrm{kg}\right) .$

Key Concepts

Unit Conversion to Electron Volts

In many quantum and atomic physics problems, energy is often expressed in electron volts (eV) instead of joules. Understanding how to convert energy values between these units is crucial for comparing theoretical predictions with experimental data.

Kinetic Energy for Massive Particles

For massive particles moving at non-relativistic speeds, their kinetic energy can be expressed as E = p²/(2m). Using the de Broglie relation where momentum p is given by h/?, one can calculate the energy associated with particles like electrons or alpha particles based on their wavelength.

de Broglie Wavelength

The de Broglie wavelength relates a particle's momentum to its wavelength through the equation ? = h/p, where h is Planck’s constant and p is the momentum. This concept is central to describing the wave behavior of both massless particles, like photons, and massive particles, such as electrons and alpha particles.

Photon Energy Relation

Photons, being particles of light, have energy directly related to their wavelength by the equation E = hc/?, where c is the speed of light. This relationship is fundamental in quantum mechanics and is widely used in calculating the energy of electromagnetic radiation.

Wave-Particle Duality

This principle explains how every particle or quantum entity exhibits both wave and particle properties. In experiments such as crystal diffraction, the wave nature of particles (including photons and matter particles) is crucial in understanding interference and diffraction patterns.

Transcript

00:01 In this exercise, we have to calculate the energy of different particles given the wavelength lambda of 0 .2 nanometers.

00:09 In question a, we have to calculate the energy of a photon.

00:15 First of all, notice that due to the debra -leave formula, given a wavelength lambda, we have that the momentum of the particle is h divided by lambda.

00:27 I'll notice that the energy of a photon is the momentum of the photon times the speed of light c.

00:35 So the energy will be hc divided by lambda.

00:39 Hc is 1 ,140 electron volts nanometers, while lambda is 0 .2 nanometers.

00:50 So this is equal to 6 ,200 electron volts.

00:55 This is the energy of a photon that has the wavelength of 0 .2 nanometers.

01:01 In question b, we have to calculate the energy of an electron.

01:07 Again, p will be equal to h over lambda.

01:11 This is valid for any particle.

01:15 The difference is that the energy of a massive particle is not equal to p times c.

01:21 Now, let's calculate what is the energy of a massive particle, a non -reactivistic massive particle.

01:30 The energy will be just the kinetic energy, and that is.

01:33 Is equal to mv squared over 2.