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quiz7soln - 3 d yz = √ 2 81 √ π ± 1 a o ² 3 2 1 a 2...

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1 Problem 1 The d orbitals have the nomenclature d z 2 , dxy , d xz , d yz , and d x 2 - y 2 . Show how the d orbital given below can be written in the form y z F ( r ). ψ 3 d yz = 2 81 π ± 1 a o ² 3 / 2 r 2 a 2 o e - r/ 3 a o sin ( θ ) cos ( θ ) sin ( φ ) 1.1 Solution z = r cos ( θ ) x = r sin ( θ ) cos ( φ ) y = r sin ( θ ) sin ( φ ) Thus, ψ 3 d yz = 2 81 π ± 1 a o ² 3 / 2 r 2 a 2 o e - r/ 3 a o sin ( θ ) cos ( θ ) sin ( φ ) ψ
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Unformatted text preview: 3 d yz = √ 2 81 √ π ± 1 a o ² 3 / 2 1 a 2 o e-r/ 3 a o r sin ( θ ) sin ( φ ) | {z } y r cos ( θ ) | {z } z ψ 3 d yz = F ( r ) y z 1...
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