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PS5Solutions - CHEM 120A Problem Set 5 Solutions David...

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Unformatted text preview: CHEM 120A Problem Set 5 Solutions David Hoffman , Doran Bennett , Tara Yacovitch 3/5/2010 1. Please see M&S Solution manual. 2. Please see M&S Solution manual. 3. We are looking for properties of the electron probability distribution for a hydrogen atom in a number of different states. To start with, recall that the hydrogen atom eigenstate is fully characterized by 3 quantum numbers ( n,l,m ). In this problem we are only interested in the radial distribution, so the quantum number m is not important. To convince yourself of this, look at the quantum states in M&S table 6, and see how the radial component compares between the (1 , 1 , 1), (1 , 1 ,- 1) and (1 , 1 , 0) states. You should find that only the θ and φ distributions are changed between states with the same n and l, but different m quantum numbers. Second, we need to remember that the correct description of a radial probability density is given by the Radial Distribution Function: P ( r ) dr = 4 πr 2 R 2 nl ( r ) dr . The reason we need this factor of r 2 is to correct for the volume associated with a given region - remember in Cartesian coordinates if you turn your space into little blocks with volume dxdydz , then all the little blocks have the same volume (you have broken your space into simple cubes). For spherical coordinates, however, if you break your space up into little regions drdθdφ then this volume is a little sliver of a spherical shell, but as r increases the size of that sliver will also increase. To correct for this effect we use the Jacobian of the system - which for spherical coordinates is given by dV = r 2 sin ( θ ) drdθdφ . So our radial probability is weighted by the additional factor of r 2 ....
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PS5Solutions - CHEM 120A Problem Set 5 Solutions David...

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