237_final_study_guide_solnsA - Physics 237 – Final Exam...

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Unformatted text preview: Physics 237 – Final Exam Review – SOLUTIONS Disclaimer: I have not seen the final exam and will not be participating in its construction. Therefore, these practice problems may not represent the actual final exam. If you don’t want to use any or some of these questions for study, that is your decision to make. Furthermore, I may not have time to write solutions. You are highly encouraged to study in groups so that you may discuss all topics thoroughly. 1. Relations a. Einstein said that E = h " and p = h " . Is this energy the kinetic, potential, rest, or what? These equations apply to photons. The rest mass of a photon is zero. The ‘free’ photon always moves at the speed of light. So these Einstein relations apply to a free photon (no potential) where the kinetic energy IS the energy of the photon. The description of light in/near material is a complicated business. Read a verbal description at the “Photon” wikipedia article, subsection: “Photons in Matter”. Note that you should be able to derive the momentum relation from the energy relation using E photon =pc. b. What did de Broglie say? Is his ‘E’ the kinetic, potential, rest, or what? de Broglie states that the same relations apply to particles. Let me first point out that for a free particle in QM we use H=p 2 /2m, i.e. only the kinetic energy is what gives wave solutions . So the de Broglie relations apply to the kinetic energy of a free particle. However, special relativity is taken into account in the actual expressions for p and E in that you must use the relativistic momentum and relativistic kinetic energy. See the wikipedia article, “de Broglie hypothesis”. (If you would like to learn more about incorporating the rest mass energy into the SWE, try reading the wikipedia article “Dirac equation”.) c. Describe the energy of a planet at rest and the energy of a planet in motion. Pretend the planet is a single particle called ‘planetonium’. Planetonium at rest does not have a de Broglie wavelength because it has zero kinetic energy and zero momentum. But planentonium in motion does have a de Broglie wavelength. For planetonium of mass 6x10 24 kg, R=6x10 6 m, linear velocity v l =3x10 4 m/s, and rotational velocity v r =5x10 2 m/s. The kinetic energy comes from linear motion and rotation. Since v<<c for both velocities, non-relativistic dynamics is valid. The rotational kinetic energy is (1/2)I ω 2 = (1/2)(2/5)Mv 2 = (1/5)Mv r 2 =3x10 29 J. The linear kinetic energy is (1/2)Mv l 2 = 3x10 33 J. Since the rotational kinetic energy is 10-4 times the size of the linear kinetic energy, we can ignore it. Therefore, ν =3x10 33 /6x10-34 =5x10 66 Hz. That’s a pretty large frequency. Meanwhile, the de Broglie wavelength of planetonium is given by l=h/p=h/[Mv]=3x10-63 m. That is a pretty small wavelength. But that’s what you get with planetonium....
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237_final_study_guide_solnsA - Physics 237 – Final Exam...

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