Section5_Compton_Effect_soln.notebook
Section 5:
May 24, 2012
The Compton Effect and Photon Momentum
In this lesson you will • Describe the Compton Effect experiment carried out by A.H. Compton • Define the term scattering • Express an understanding of Compton's pioneering view that a photon of light has momentum and hence the properties of mass
• Use Einstein's mass­energy equivalency expression and Planck's equation to derive an expression for the momentum of a photon
• Express an understanding for Compton's role in showing that light
possesses both wave and particle properties at the fundamental
level • Use the mathematical expressions in this lesson to solve numerical exercises 1
Section5_Compton_Effect_soln.notebook
May 24, 2012
In order to discuss the Compton Effect and Photon Momentum, you have to have an understanding of:
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the law of conservation of energy the law of conservation of momentum Planck's equation Einstein's mass­energy equivalency expression Einstein’s Mass­Energy Equivalency expression:
Sometimes it is difficult to tell if light is a particle or a wave. Light behaves as a wave when it interferes constructively and destructively to produce interference patterns (Page 427 of text), but it behaves as a particle when is causes the photoelectric effect. Moreover, light particles , or photons, have wave characteristics ­­for example, the energy of 1 photon is hf, and "f" is definitely a wave characteristic. As crazy as it may sound, light is a wave is a particle!
Einstein's mass­energy equivalency expression is depicted by Einstein's most famous equation, E = mc2. When arranged as m = E / c2, the term E / c2 is called the mass equivalency .
E = mc2 makes a statement that you may find difficult to accept. It says that mass is energy, and energy is mass. Mass and energy are one and the same . It is difficult to accept this when you look at a baseball, or a house, or your best friend. It may be a little easier to accept when if you consider interactions at an atomic level, rather than at an every day level. Even so, you may be wondering why we don't have more energy than we can handle. For example, if you burn a 1.0 kg birch junk in your fireplace, how is it that you don't heat up the whole province? After all
E = mc 2 = 1.0 kg x 3.0 x 10 8 m/s x 3.0 x 10 8 m/s
= 9.0 x 10 16 J !!!!!!!! 2
Section5_Compton_Effect_soln.notebook
May 24, 2012
Answer: If, while the junk was burning you could catch all the gases and smoke particles that went up the chimney and added the mass of the gases and smoke to the mass of the ash left behind, your scale would give a reading of 1.0 kg. But be careful! You can't conclude that no mass was changed into energy . All you can say is the amount of mass "lost" is so small that your scale is not sensitive enough to detect it. Say, for the sake of this discussion that the amount of birch that provides all that nice heat in your room is in the approximate order of 1 part in 1,000,000,000,000. That is, say approximately one trillionth of the junk (that's 10­12 kg) is actually converted into energy!! Now E = mc2 makes a little more sense:
E = mc2 = 10 ­12 kg x (3.0 x 10 8 m/s)2 = 9.0 x 10 9.0 x 10 4 J
By now you are probably saying, "So, what the big deal about E=mc2?"
The big deal is this: your burning birch junk is a chemical reaction­­it is chemical combustion . In chemical combustion an immeasurable amount of the original mass gets converted into energy. However, in a nuclear reaction the proportion of mass that can be changed into energy is not 1 part in 1,000,000,000,000, but rather is more like 1 part in 1000!! When such a nuclear reaction takes place in a reactor, this is good. When such a reaction takes place in an atomic bomb, this is bad!!!
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Section5_Compton_Effect_soln.notebook
May 24, 2012
The Compton Effect and Photon Momentum (Textbook: Section 17.4, pp. 707­710.)
Arthur Compton carried out experiments to determine how photons (light particles) react with matter. Photons were defined as being entirely energy and having no mass.
Compton designed and carried out an experiment in which x­rays of photons were directed at a thin piece of foil containing a target made of carbon. He arranged detectors behind the carbon, and found that once the x­rays hit the carbon, it emitted other x­rays and some electrons . This phenomenon was referred to as scattering because the x­rays are deflected in many different directions He discovered that the second lot of x­rays had lower energy than the incident x­rays . Compton found that the energies of the scattered photons were very different from those of the incident x­ray photons. This was difficult to explain since photons were entirely energy having no mass, and theoretically, they should not experience any changes to their energy during collisions.
The picture above resembles a collision of pool balls on a pool table. Compton must have noticed this too because he decided to treat photons as particles that collided with electrons. Thus, he reasoned that they should obey the laws of conservation of energy and momentum. 4
Section5_Compton_Effect_soln.notebook
May 24, 2012
Law of Conservation of Energy:
The energy of the incoming x­ray is divided between the scattered x­
rays and the electron: Ex­ray = E scattered + E electron
but E = hf and Eelectron = ½ mv2
hfx­ray = hf scattered + ½ m ev2.
So,
Law of Conservation of Momentum :
The momentum of the incoming photon is divided between the scattered x­rays and the electron:
Rearranging E = mc2 for mass gives m = E/c2. So,
p = mv
but m = E/c 2 and for photons v = c
but
where: h is plank’s constant and λ is the wavelength in meters Conclusions from Compton’s Work :
1. Photons collide and exchange energy with particles according to the law of conservation of energy.
2. A photon does indeed have momentum. The momentum depends on wavelength and Planck’s constant according to the equation
3. The momentum of a photon is conserved during collisions.
His work lent support to the idea that light possessed both wave and particle properties at a fundamental level. It indicated that investigations of these phenomena were at the interface between what is considered matter and what is considered energy.
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Section5_Compton_Effect_soln.notebook
May 24, 2012
Examples
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In a Compton experiment, a 82 eV x­ray photon collides with an electron causing the scattering of a lower energy x­ray of wavelength 195 nm. The mass of an electron is 9.11 x 10 ­31 kg.
Determine
a) the momentum of the original x­ray.
b) the momentum of the scattered x­ray.
c) the energy imparted to the electron.
d) the increase in speed of the electron.
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Section5_Compton_Effect_soln.notebook
2
May 24, 2012
A scattered x­ray photon has 75% of the momentum of the incident photon. If the wavelength of the incident photon is 150 nm, what is the wavelength of the scattered photon?
Practice
1. In the Compton Effect let E1 be the energy of the incoming photon and E2 be the energy of the emitted x­ray. Which statement is true?
a) E1 > E2 b) E1 is sometimes > and sometimes < E2 ,
c) E1 < E2 d) E1 = E2
2. Which of the following expressions is not a correct one for the momentum of a photon? a) p = h/λ, b) p = hc/f c) p = E/c, d) p = hf/c 7
Section5_Compton_Effect_soln.notebook
May 24, 2012
3. When a photon is scattered from an electron, which of the following properties of the photon will increase? a) its energy c) its wavelength b) its frequency d) its momentum 4. What is the momentum of a photon that has a wavelength of 2.5 x 10 2 nm? a) 2.7 x 10 ­36 Ns c) 1.2 x 10 15 Ns b) 1.2 x 10 6 Ns d) 2.7 x 10 ­27 Ns 5. The energy of a 150 eV photon diminishes by 25% after scattering. What is the wavelength of the emitted (scattered) photon? a) 11 nm c) 1.1 x 10 ­7 nm b) 18 nm d) 1.8 x 10 ­7 6. Calculate the magnitude of the momentum of a photon with wavelength 625 nm. 7. A photon has a frequency of 7.8 x 10 16 Hz. What is its momentum? 8
Section5_Compton_Effect_soln.notebook
May 24, 2012
8. What is the momentum of a photon whose energy is 225 eV? 9. A scattered photon has the same momentum as that of an emitted electron that travels at 8.8 x 10 5 m/s. What is frequency and wavelength of the photon?
In your textbook: on p. 710­­do #2. on p. 732­­do #30, #32
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