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Unformatted text preview: 5.61 Physical Chemistry Lecture #26 page 1 TWO ELECTRONS: EXCITED STATES In the last lecture, we learned that the independent particle model gives a reasonable description of the ground state energy of the Helium atom. Before moving on to talk about manyelectron atoms, it is important to point out that we can describe many more properties of the system using the same type of approximation. By using the same independent particle prescription we can come up with wavefunctions for excited states and determine their energies, their dependence on electron spin, etc. by examining the wavefunctions themselves. That is to say, there is much we can determine from simply looking at Ψ without doing any significant computation. We will use the excited state 1s2s configuration of Helium as an example. For the ground state we had: Ψ space ( r 1 , r 2 ) × Ψ spin ( σ 1 , σ 2 ) α σ β σ − β σ α σ ⇒ ψ ( r ) ψ ( r ) 1 2 ⎜ ⎝ ⎛ ( 1 ) ( 2 ) ( 1 ) ( 2 ) ⎟ ⎠ ⎞ 1 s 1 1 s 2 1s In constructing excited states it is useful to extend the stick diagrams we have used before to describe electronic configurations. Then there are four different configurations we can come up with: Ψ space ( r 1 , r 2 ) × Ψ spin ( σ 1 , σ 2 ) 2s ? ? 1s 2s ? ? 1s 2s ? ? 1s 2s ? ? 1s 5.61 Physical Chemistry Lecture #26 page 2 Where the question marks indicate that we need to determine the space and spin wavefunctions that correspond to these stick diagrams. It is fairly easy to come up with a reasonable guess for each configuration. For example, in the first case we might write down a wavefunction like: ψ r σ ψ r σ = ψ r ψ r α σ α σ 1 s α ( 1 1 ) 2 s α ( 2 2 ) 1 s ( 1 ) 2 s ( 2 ) ( 1 ) ( 2 ) However, we immediately note that this wavefunction is not antisymmetric. We can perform the same trick as before to make an antisymmetric wavefunction out this: 1 ⎛ ψ r ψ r α σ α σ − ψ r ψ r α σ α σ ⎞ ⇒ ⎜ 1 s ( 1 ) 2 s ( 2 ) ( 1 ) ( 2 ) 1 s ( 2 ) 2 s ( 1 ) ( 2 ) ( 1 ) ⎟ 2 ⎝ ⎠ ⇒ 1 ⎛ ψ r ψ r − ψ r ψ r ⎞ α σ α σ ⎜ 1 ( 1 ) 2 s ( 2 ) 1 s ( 2 ) 2 s ( 1 ) ⎟ ( 1 ) ( 2 ) 2 ⎝ s ⎠ Ψ Ψ space spin Applying the same principle to the 1s ↑ 2s ↓ configuration gives us a bit of trouble: ⇒ ⎜ ⎛ ⎝ ψ ) ) β σ ) ) ( ( ⎞ ⎠ Ψ space Ψ ψ 1 s ( r 1 ) 2 s ( r 2 α σ ( 1 ( 2 − ψ 1 s ( r 2 ψ 2 s ( r 1 ) α σ 2 ) β σ 1 ) ⎟ ≠ spin Hence, the pure ↑↓ configuration can’t be separated in terms of a space part and a spin part. We find a similar result for 1s ↓ 2s ↑ : ⇒ 1 ψ r ψ r β σ α σ − ψ r ψ r β σ α σ ≠ Ψ Ψ space 2 ⎜ ⎝ ⎛ 1 s ( 1 ) 2 s ( 2 ) ( 1 ) ( 2 ) 1 s ( 2 ) 2 s ( 1 ) ( 2 ) ( 1 ) ⎟ ⎠ ⎞ spin Since we know the wavefunction should separate, we have a problem. The solution comes in realizing that for an open shell configuration like this one, the 1s ↑ 2s ↓ and 1s ↓ 2s ↑ states are degenerate eigenstates and so we can make any linear combinations of them we like and...
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 Spring '08
 greenwood
 Calculus

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