Lecture9ForcesPotentialsandtheShellModel

Lecture9ForcesPotentialsandtheShellModel - Forces,...

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Forces, Potentials, and the Shell Model Recall the Infinite Square Well (1D) Solve Shroedinger’s equation: ψ E H = E V dx d = - 2 2 Result: Consideration of boundary conditions (the behavior of the wavefunction at the walls) results in quantization. Both wavefunctions and eigenstates (energy levels) 2 2 2 8 mL h n E n = Notice the dependence of the energy levels on the size of the box, and on the principal quantum number. Harmonic oscillator (1D) Hooke’s law : ) ( 0 x x k F - - = If 0 x x = , the system is at equilibrium because there is no force. However if x is different from 0 x there is a force which acts to restore the position to the equilibrium value (Notice the negative sign.) dx dV F - = Integrating we get, 2 0 ) ( 2 1 x x k V - = Now solve Schrodinger’s equation using this potential. Solution: Wavefunctions and eigenvalues Eigenvalues: ϖ ) 2 1 ( + = n E n where m k =
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Notice the energy spacing for the harmonic oscillator. What is the minimum energy of the harmonic oscillator?
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V. Nuclear Shell Model A. Quantum Properties of Nuclei 1. Discrete Energy Levels 2. Nuclear Spin - I a. Experimental Summary e-e : 0 = I ALWAYS e-o, o-e: 2 e n I = , where n is an odd integer (1/2, 3/2, . ..) o-o : e n I = , where n is an integer (0, 1, 2 . ..) WE'LL USE 1 = for our spins b. Implication e-e result implies strong pairing is energetically favorable spins must cancel c. Reason: Nuclear Force is attractive ; in contrast spins are unpaired in a atomic orbitals due to e-e repulsion ( Pauli exclusion principle ) 3. Closed Shells – Unusual Stability a. Magic Numbers 2, 8, 20, 28, 50, 82, 126 (neutrons) b. Energetics: (M LD – M), B p , B n , B α c. Lifetimes: 8 2 2 0 8 1 2 6 P b 8 2 2 0 9 1 2 7 P b 8 4 2 1 0 P o 1 2 6 8 4 2 1 2 1 2 0 P o STABLE 22y 138d 10 - 7 s Z=82 & N=126 appear to be stable
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4. Magnetic Moments Moving Charge created a magnetic field with moment μ μ = e 2 M c f ( I ) e ; μ N = nuclear magneton (M = M p ) a. Expect μ p = μ N Observe : μ p = 2.793 μ N μ n = 0 μ n = - 1.913 μ N b. Implication: nucleon has substructure, since one observes charge on periphery of particle. e.g., proton
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This note was uploaded on 07/02/2011 for the course CHEM-C 460 taught by Professor Staff during the Spring '10 term at Indiana.

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Lecture9ForcesPotentialsandtheShellModel - Forces,...

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