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Unformatted text preview: Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2010 Laura Gagliardi Lecture 3, September 13, 2010 Solved Homework The total power emitted by a blackbody is obtained by integrating the frequency- dependent energy density over all frequencies. Thus, we must solve T ( ) = 8 h 3 c 3 e h / kT 1 ( ) d To solve this integral, it is most convenient to introduce a new variable x = h / kT . In that case, we also have dx = ( h / kT ) d . We may rearrange these to see what substitutions we will make in the integral for and d , namely = xkT / h and d = ( kT / h ) dx . If we introduce these substitutions we have T ( ) = 8 h kTx h 3 c 3 e x 1 ( ) kT h dx = 8 k 4 T 4 c 3 h 3 x 3 e x 1 ( ) dx The value of the final integral can be looked up: it is 4 /15. If one plugs in all the appropriate values for the various constants, one obtains = aT 4 where a = 7.5657 x 10 16 J m-3 K-4 . This derivation from Planck's formula agrees essentially perfectly with experiment. Page 1-25 of the Book (Chapter 1) The Orbiting Electron Model of the Hydrogen Atom In the first decade of the 1900s, it began to become clear that atomic structure consisted of massive nuclei, composed of protons and neutrons, surrounded by electrons that had comparatively enormous volumes of empty space available to them. Protons and electrons attract one another through the Coulomb force (the only one of the four physical forces to really matter in Chemistry). Since atoms do not spontaneously annihilate their charged particles in fractions of a second, there must be an opposing force that repels the electrons from the nucleus. The situation just described has an obvious analogy (particularly to physicists), namely the centrifugal force of an orbiting body that opposes the gravitational force attracting it to a central mass, e.g., a planet. 3-2 The total energy of a "planetary" hydrogen atom, assuming the nucleus to be stationary, would derive from two terms: the kinetic energy of the orbiting electron and the potential energy from the Coulomb attraction. We will address each term individually. Kinetic Energy for an Orbiting Body A convenient expression for kinetic energy, usually denoted T (not to be confused with temperature!), is T = p 2 2 m (3-1) where p is momentum (mass times velocity, units of mass-distance per timeit is a trivial matter to verify that this agrees with the other common expression T = (1/2) mv 2 ) (3-2) For an orbiting body, however, it is usually more convenient not to discuss mass, velocity, and momentum, but rather the analogous quantities, moment of inertia, angular velocity, and angular momentum. To see the relationship between them easily, it is helpful to visualize the system....
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