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Unformatted text preview: Prob 29.1 – A and C Electron in state 5f, so angular momentum quantum number l = 3, magnetic quantum number m = +/ l = +/3, 2, 1, or 0 L = l l 1 ℏ L = 3 3 1 ℏ L = 2 3 ℏ Prob 29.2 – A and D 17 electrons – add up superscripts in configuration there is an electron in the p state, which corresponds to l = 1, so m = +/1 or 0 Prob 29.3 – B and D Orbital Angular Momentum for Hydrogen, Part A – L = 2.45hbar n = 3 l = n – 1 = 2 L = l l 1 ℏ L = 2 2 1 ℏ L = 2.45 ℏ Part B – L = 32.5hbar n = 33 l = n – 1 = 32 L = l l 1 ℏ L = 32 32 1 ℏ L = 32.5 ℏ Part C L = 212hbar n = 213 l = n – 1 = 212 L = l l 1 ℏ L = 212 212 1 ℏ L = 212 ℏ Rules for Orbital Angular Momentum, Part A – 4 n = 4 l = 3, 2 1, 0 Part B – 3 l = 1 m = 1, 0, 1 Part C – 18 Part D No: The orbital quantum number cannot equal the principal quantum number....
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This note was uploaded on 06/07/2008 for the course PHYS 6C taught by Professor Geller during the Spring '08 term at UCSB.
 Spring '08
 Geller
 Angular Momentum, Momentum

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