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# HW3S - Gauthier Joseph Homework 3 Due midnight Inst...

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Gauthier, Joseph – Homework 3 – Due: Sep 18 2007, midnight – Inst: Vandenbout 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. This homework is the second of two home- works from Chapter 12 in your textbook. 001 (part 1 of 1) 10 points What is the de Broglie wavelength of a flea midway through its jump? Assume the mass of the flea is 4 . 50 × 10 - 4 g and it is moving at 1.00 m/s. 1. 1 . 47244 × 10 - 27 m correct 2. 2 . 9818 × 10 - 37 m 3. 2 . 9818 × 10 - 40 m 4. 1 . 4725 × 10 - 30 m Explanation: m = 0 . 00045 g × 0 . 001 kg 1 g = 4 . 5 × 10 - 7 kg λ = h p = h m · v = 6 . 626 × 10 - 34 kg · m 2 s (4 . 5 × 10 - 7 kg)(1 m / s) = 1 . 47244 × 10 - 27 m 002 (part 1 of 1) 10 points An electron in a 3 d orbital could have which of the following quantum numbers? 1. n = 3; = 0; m = 0 2. n = 3; = 1; m = - 1 3. n = 3; = 1; m = 2 4. n = 3; = 3; m = 1 5. n = 3; = 2; m = 0 correct 6. n = 2; = 2; m = 6 7. n = 3; = 2; m = - 3 Explanation: 3 refers to the principal quantum number n . d corresponds to the subsidiary quantum number = 2. Since = 2, m could be - 2 , - 1 , 0 , 1 or 2. 003 (part 1 of 1) 10 points How many p electrons does Se (atomic num- ber 34) possess? 1. 16 correct 2. 0 3. 6 4. 34 5. 10 6. 12 7. 4 Explanation: Se has 6 electrons in the 2 p and 3 p orbitals plus 4 more in the 4 p orbital. This gives it a total of 16 p electrons. 004 (part 1 of 3) 10 points What is the principal quantum number for the orbital 5 f ? 1. 5 correct 2. 2 3. 6 4. 4 5. None of these 6. 3 Explanation:

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Gauthier, Joseph – Homework 3 – Due: Sep 18 2007, midnight – Inst: Vandenbout 2 n = 5 005 (part 2 of 3) 10 points What is the orbital angular momentum quan- tum number for the orbital 5 f ? 1. 1 2. 3 correct 3. None of these 4. 2 5. 4 6. 0 Explanation: = 3 006 (part 3 of 3) 10 points How many magnetic quantum numbers does the orbital 5 f have? 1. None of these 2. 3 3. 9 4. 5 5. 7 correct 6. 1 Explanation: m = - 3 , - 2 , - 1 , 0 , 1 , 2 , 3 007 (part 1 of 1) 10 points How many total nodal planes are present in the complete set of 3 d orbitals?
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HW3S - Gauthier Joseph Homework 3 Due midnight Inst...

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