Physics 30 - Lesson 23： The Compton Effect and the de Broglie Wavelength

Lesson 25. The Compton Effect and the de Broglie Wavelength. Part 1. The Compton Effect. Arthur Compton was intrigued by the idea of light being made up of particles of energy. Since particles have momentum, he wondered if light could have momentum too. The momentum of a particle with mass, or we call this a massive particle, is given by p equals mv. Einstein showed that energy and mass are equivalent in his energy-mass equivalence equation, E equals mc squared. We can rearrange this equation for m. m equals E over c squared. So mass is equivalent to energy. And since the speed of light is 3 times 10 to the 8 meters per second, we can take our momentum formula, p equals mv, and substitute in the speed of light and the mass-energy equivalence to get p equals E over c squared times c. But since c appears twice in this equation, it can be cancelled out. This leaves us with p equals E over c. But we're not quite done yet. Since the energy of the photon is E equals hf, Compton predicted the momentum of a photon would be p equals E over c. So p equals hf over c. This is Compton's equation to calculate the momentum of a photon. We also know that for a photon, the wavelength is given by c over frequency. So we could also calculate the momentum of a photon using momentum equals h over lambda. For example, calculate the momentum of a photon with a wavelength of 450 nanometers. We can use the formula p equals h over lambda to calculate this momentum. p is equal to h, which is 6.63 times 10 to negative 34, that's Planck's constant, divided by the wavelength, which is 450 nanometers, which is 450 times 10 to the negative 9, since nano is negative 9. This gives a momentum equal to 1.47 times 10 to the negative 27. And the unit for this, even though the photon doesn't have mass, is still the standard momentum unit, kilogram meters per second. To test his theory, Compton bombarded electrons with photons and measured the velocity of the electrons after the collision. It was a standard hit and bounce collision. There are three ways you can analyze these collisions. The first is conservation of energy. The energy before the collision is equal to the energy after the collision. The second is conservation of momentum. The total momentum before the collision is equal to the total momentum after. And the third is an equation Compton developed himself. More on that later. We'll start with conservation of energy. He measured the energy gain of the electron and found it to be equal to the energy loss of the photon. The total energy before was equal to the total energy after. The total energy before was the energy of the photon plus the energy of the electron before the collision. The total energy after was the energy of the photon after the collision plus the energy of the electron after the collision. The energy of a photon is given by hf. So the energy of the photon before the collision was hf before. The energy of an electron is usually only its kinetic energy. This is 1 half mv before the collision squared. The energy of the photon after is hf after, and the energy of the electron is 1 half mv after squared. The photon can't change speed. The speed of light is fixed. So in order to lose energy, the frequency must change. It has a different frequency before the collision and after the collision. Next, let's look at conservation of momentum. The total momentum before the collision equals the total momentum after the collision. The momentum of the photon plus the momentum of the electron before the collision equals the momentum of the photon plus the momentum of the electron after the collision. But it's more complicated than energy, because momentum is a vector. This means direction and angle matters in our calculations. We can draw a horizontal line down the center of the collision, so we can measure the angle of each particle before and after the collision. Theta of the photon before the collision, theta of the photon after the collision, theta of the electron before the collision, and theta of the electron after the collision. Momentum is conserved in each direction separately, so we have to find the components of the momentum in both the x and y directions. Let's look at the momentum of the photon after the collision. It has a y and x component. Since the y component is opposite to the angle, we can use sine. Momentum in the y direction is equal to momentum of the photon sine theta. Since the x component is adjacent to the angle, we can use cos to find the x component. Momentum in the x direction equals momentum of the photon cos theta. So in the x direction, we have momentum of the photon cos theta photon before plus momentum of the electron cos theta electron before equals momentum of the photon cos theta photon after plus momentum of the electron cos theta electron after. In the y direction, we have momentum of the photon sine theta photon before plus momentum of the electron sine theta electron before equals momentum of the photon sine theta photon after plus momentum of the electron sine theta electron after. The photon is not a massive particle. To find the magnitude of the momentum of the photon, we use h over lambda before in our equations. The electron is a massive particle, so to find the momentum of the electron, we use p equals mv. After the collision, the photon is h over lambda after, and the electron is mv after. We substitute our knowns into each of these equations, and then we solve for the unknown. Once we have the unknown value in the x and y direction, we can put them back together using vector addition. The third method to analyze these electron photon collisions is using the Compton equation. Using both conservation of energy and conservation momentum, along with the application of Einstein's special theory of relativity, Compton derived the following relationship for a change in wavelength of the x-ray photon. Delta wavelength is equal to h over mc 1 minus cos theta, where m is the mass of the electron, and theta is the angle through which the x-ray photon scatters. Remember, delta lambda is the change in wavelength. This is final wavelength, minus initial wavelength. So we have three methods to analyze the Compton effect. The first is conservation of energy. Conservation of energy is best for calculating the final speed of the electron. The second method, conservation of momentum, you want to avoid if it is all possible, because it's the hardest method to use because of the angles. It's best for calculating the final velocity or angle of the electron. The third method, the Compton equation, is best used for calculating the wavelength or the angle of the photon. Example, a photon of light with a wavelength of 0.427 nanometers hits a stationary electron. The scattered x-ray has a wavelength of 0.429 nanometers. What is the resulting speed of the electron and the scattering angle of the x-ray? This question asks for two things, the resulting speed of the electron and the scattering angle of the x-ray. The speed of the electron is best found using the first method, conservation of energy. The scattering angle of the x-ray photon is best solved for using the third method, the Compton equation. We'll start by finding the speed of the electron. This is, uses conservation of energy. The sum of the energies before equals the sum of the energies after. Before, we have the photon hc over lambda before, plus the energy of the electron, one-half mv before squared, equals the energy of the photon after, hc over lambda after, plus the energy of the electron after, one-half mv after squared. The wavelength before the collision was 0.427 nanometers. The electron was originally stationary, which means it was going 0 meters per second before the collision, which means the initial kinetic energy of the electron before the collision is zero. The wavelength after the collision is 0.429 nanometers, and the velocity after the collision is our unknown that we're looking for. We can rearrange this for the velocity of the electron after. One-half mv after squared equals hc over lambda before, minus hc over lambda after. This is one-half mv after squared equals hc over lambda before as Planck's constant times the speed of light divided by the wavelength before the collision. This is equal to 4.6581 times 10 to the negative 16, minus hc over lambda after, which is 4.6363 times 10 to the negative 16. This gives us one-half mv after squared equals a change in energy equal to 2.17 times 10 to the negative 16 joules. To solve for velocity, we multiply this result by 2 and divide by m square rooting. This gives us a velocity after equal to 2.18 times 10 to the 7 meters per second. Next we're going to use the Compton equation to determine the scattering angle of the x-ray after the collision. The Compton equation says the change in wavelength equals h over mc, 1 minus cos theta. We can rearrange this for theta. First we'll move mc and h. Delta lambda times mc divided by h equals 1 minus cos theta. Next we isolate the cos theta. Cos theta equals 1 minus change in wavelength mc over h. We can solve for theta by inverse cos-ing everything on the other side of the equation. So theta equals cos negative 1, 1 minus change in wavelength mc over h. Since the change in wavelength is theta final minus theta initial, theta equals cos negative 1, 1 minus theta final minus theta initial mc over h. This means theta equals cos negative 1, 1 minus 0.429 times 10 to the negative 9 minus 0.427 times 10 to the negative 9 times the mass of an electron which is 9.11 times 10 to the negative 31 times the speed of light which is 3 times 10 to the 8 divided by Planck's constant which is 6.63 times 10 to the negative 34. This gives us theta equals 79.9 degrees. Here's another example. An x-ray photon of wavelength 0.0500 nanometers scatters at an angle of 30 degrees. Calculate the wavelength of the photon after the collision. We start by finding the change in wavelength. This is done using Compton's formula. Delta lambda equals h over mc 1 minus cos theta. The change in wavelength equals 6.63 times 10 to the negative 34, that's Planck's constant, divided by the mass of the electron which is 9.11 times 10 to the minus 31 times the speed of light which is 3 times 10 to the 8 multiplied by 1 minus cos, the angle which is 30 degrees. This gives us a change in wavelength equal to 3.25 times 10 to the negative 13 meters. Since the change in wavelength is equal to lambda final minus lambda initial, therefore the final wavelength is equal to the change in lambda plus lambda initial. So theta final is equal to 3.25 times 10 to the negative 13 plus 0.05 times 10 to the minus 9. This gives us a final wavelength equal to 0.0503 nanometers. Part 2, the de Broglie wavelength. In the 1800s, light was proven to act like a wave. It reflects, refracts, diffracts, and experiences interference. Planck, Einstein, Bohr, and Compton showed that light acts like particles that comes in packets of energy, E equals hf, the energy of a photon, and has momentum, momentum equals h over wavelength. Light has wave-particle duality. Lewis de Broglie wondered if the converse was also true. Could particles with mass act like waves? De Broglie developed a formula for the wavelength of a particle. Since we know the momentum of a particle with mass is p equals mv, and we know that the momentum of a particle with a wavelength is p equals h over lambda. If particles with mass also had a wavelength, then these momenta should be equivalent. mv equals h over lambda. Rearranging this for wavelength gives us lambda equals h over mv, the wavelength of a particle with mass. We can use this to calculate the electron. Example, if an electron accelerates from rest through 100 volts, what is its de Broglie wavelength? We can use this voltage to find out the velocity of the electron. Voltage is equal to change in energy, in this case kinetic energy, divided by charge. So, the change in kinetic energy is equal to qv, and kinetic energy is 1 half mv squared. This equals qv, and we can rearrange for velocity. v equals 2qv divided by m, square rooted. This gives us the velocity of the electron equals 2 times the charge of an electron, 1.6 times 10 to the negative 19 Coulombs, times the voltage it's accelerating through, which is 100 volts, divided by the mass of the electron, which is 9.11 times 10 to the negative 31. All of this is square rooted. This gives us the velocity of the electron after it's accelerated, equal to 5.927 times 10 to the 6 meters per second. We can use this velocity to calculate the electron's de Broglie wavelength. Wavelength is equal to h over mv. This is 6.63 times 10 to the negative 34, divided by the mass of the electron, which is 9.11 times 10 to the negative 31, times 5.927 times 10 to the 6, which is its velocity. This gives us a wavelength equal to 0.123 nanometers. It's tiny, but it's not that small. X-rays have wavelengths similar to this size. Although the de Broglie wavelength for an electron is very small, the spaces between the atoms and crystals are similarly small. Two scientists, Davidson and Germer, shot electrons at nickel crystal and observed a diffraction interference pattern. This indicated that electrons can behave exactly like waves. You pass them through spaces, and they'll interfere with each other, creating a fringe pattern, just like light. Electrons, particles with mass, show wave-particle duality, just like de Broglie predicted.