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Lecture23

# Lecture23 - CHEM 356 Lecture 23 Fall 2009 1 Probabilistic...

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Unformatted text preview: CHEM 356, Lecture 23, Fall 2009 1 Probabilistic Interpretation for Hydrogenic Systems. We begin by noting that [ ℓ ( ? )] 2 ? 2 ? represents the probability that the electron for a hydrogenic system can be found at a distance from the nucleus that lies between ? and ? + ? , so that plots of ? 2 2 ℓ ( ? ) represent radial probability densities. Let us consider, for example, the (1 ? )–orbital, for which the wavefunction 1 ? ( ?, , ) is 1 ? ( ?, , ) = 1 √ 3 e − ?/ , with the Bohr radius. 1. Probability density: at any point on a sphere of radius ? the probability density is given by ( ?, , ) = ∣ 1 ? ( ? ) ∣ 2 = 1 3 e − 2 ?/ . Note that the probability density for finding such an electron at the origin of our relative coordinate system (i.e., at the location of the centre–of–mass of our electron–nucleus system) is large and finite, being ∣ 1 ? (0) ∣ 2 = 1 3 = 2 . 148 × 10 30 m − 3 = 2 . 148 ˚ − 3 2. Probability for finding the electron in a volume element = ? 2 sin ? centred on the point ( ?, , ) is given by ( ?, , ) = ∣ 1 ? ( ?, , ) ∣ 2 = 1 3 e − 2 ?/ ? 2 sin ? . 3. Probability for finding the electron at a distance ? from the electron–nucleus centre–of–mass is obtained by integrating ∣ 1 ? ( ?, , ) ∣ 2 over and : this gives ( ? ) = 1 3 e − 2 ?/ ? 2 ? ∫ sin ∫ 2 = 4 3 e − 2 ?/ ? 2 ? CHEM 356, Lecture 23, Fall 2009 2 4. Probability for finding the electron within a sphere of radius about the electron–nucleus centre–of–mass is given by (1 ? ) = 4 3 ∫ e − 2 ?/ ?...
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Lecture23 - CHEM 356 Lecture 23 Fall 2009 1 Probabilistic...

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