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Unformatted text preview: Physical Chemistry Course Number: C362 12 The Hydrogen Atom 1. As we noted in the BornOppenheimer approximation, solving the time independent molecular Schr¨odinger Equation: H Ψ( r i , R I ) = E Ψ( r i , R I ) (12.1) where H = − ¯ h 2 2 N summationdisplay I =1 1 M I ∇ 2 I − ¯ h 2 2 m n summationdisplay i =1 ∇ 2 i − N summationdisplay I =1 n summationdisplay i =1 Ze 2 r i,I + N summationdisplay I =1 N summationdisplay J =1 ′ Z I Z J e 2 r I,J + n summationdisplay i =1 n summationdisplay j =1 ′ e 2 r i,j (12.2) and φ ( r i , R I ) = ψ ( r i , { R I } ) χ ( R I ) (12.3) is not possible to do exactly for more than one electron. Approximations must be made. However, it is good to know that we can exactly solve the problem if the system contained only one electron. In fact, the solution to the problem with only one electron completely explains and is responsible for the way the periodic table is arranged. We will now look at this solution carefully. 2. There are two particles in the system, an electron and a nucleus, and so we can rewrite the Hamiltonian as: H ( vector r, vector R ) = − ¯ h 2 2 m ∇ 2 r − ¯ h 2 2 M ∇ 2 R − Ze 2 vextendsingle vextendsingle vextendsingle vextendsingle vector r − vector R vextendsingle vextendsingle vextendsingle vextendsingle (12.4) Realize that this is obtained from Eq. ( 12.2 ) above, by using N = n = 1 , that is one electron and one nucleus. Hence no internuclear repulsion and no electronelectron repulsion. 3. We have already seen earlier that the spherical coordinate system is ideal to solve such problems, on account of the spherical symmetry of atoms. [We saw something similar in the rotational spectroscopy problem. How is that relevant here? Homework ] Chemistry, Indiana University 63 c circlecopyrt 2011, Srinivasan S. Iyengar (instructor) Physical Chemistry Course Number: C362 4. There are two particles, the electron and the nucleus. Lets assume the vector vector R represents the position of the nucleus (mass M ) and the let vector r represent the position of the electron (mass m ). 5. What is the center of mass of this system? The position of the center of mass is given by the vector: vector R CM ≡ mvector r + M vector R m + M (12.5) and lets introduce the vector: vector r e − N ≡ vector R − vector r (12.6) Or we can write the old coordinates in terms of the new as: vector R = vector R CM + m m + M vector r e − N (12.7) and vector r = vector R CM − M m + M vector r e − N (12.8) Homework: Show Eqs. ( 12.7 ) and ( 12.8 ) to be true....
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This note was uploaded on 01/17/2012 for the course C 362 taught by Professor Amarflood during the Winter '11 term at Indiana.
 Winter '11
 AmarFlood

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