McCord - 2010 - Exam 1-solutions

McCord - 2010 - Exam 1-solutions - Version 238 Exam 1...

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Unformatted text preview: Version 238 Exam 1 McCord (53110) 1 This print-out should have 29 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. McCord CH301mwf This exam is only for McCords MWF CH301 class. c = 3 . 00 10 8 m/s h = 6 . 626 10- 34 J s m e = 9 . 11 10- 31 kg N A = 6 . 022 10 23 mol- 1 = R parenleftbigg 1 n 2 y- 1 n 2 x parenrightbigg where R = 3 . 29 10 15 s- 1 n ( x ) = parenleftbigg 2 L parenrightbigg 1 2 sin parenleftBig nx L parenrightBig E n = n 2 h 2 8 mL 2 n = 1 , 2 , 3 , 001 10.0 points Which of the following would be expected to have the lowest first ionization energy? 1. Li correct 2. F 3. Ne 4. O 5. C Explanation: Ionization energy generally decreases from right to left across the Periodic Table. 002 10.0 points If a particle is confined to a one-dimensional box of length 300 pm, for 3 the particle has zero probability of being found at 1. 50, 150, and 250 pm, respectively. 2. 50 and 250 pm, respectively. 3. 75, 125, 175, and 225 pm, respectively. 4. 100 and 200 pm, respectively. correct 5. 150 pm only. Explanation: 100 200 300 003 10.0 points A quantum mechanical particle in a box in its ground state is most likely to be found 1. is equally likely to be found at all positions except the very center of the box. 2. at the very edge of the box. 3. is equally likely to be found at all positions in the box. 4. in the middle of the box. correct 5. at both edges of the box. Explanation: The ground state is 1 which has maximum probability in the middle of the box. 004 10.0 points In the spectrum of atomic hydrogen, a blue line is observed at 486 nm. What are the be- ginning and ending energy levels of the elec- tron during the emission of energy that leads to this spectral line? 1. n = 6, n = 3 2. n = 3, n = 2 3. n = 4, n = 3 Version 238 Exam 1 McCord (53110) 2 4. n = 5, n = 2 5. n = 6, n = 2 6. n = 5, n = 3 7. n = 4, n = 2 correct Explanation: = 486 nm = 4 . 86 10- 7 m Because the line is in the visible part of the spectrum, it belongs to the Balmer series for which the ending n is 2. For the starting value of n , = c = 3 10 8 m / s 4 . 86 10- 7 m = 6 . 16875 10 14 s- 1 Using the Ryberg formula, = (3 . 29 10 15 s- 1 ) parenleftbigg 1 n 2 2- 1 n 2 1 parenrightbigg 3 . 29 10 15 s- 1 = parenleftbigg 1 n 2 1- 1 n 2 2 parenrightbigg 1 n 2 2 = 1 n 2 1- 3 . 29 10 15 s- 1 = 1 4- 6 . 16875 10 14 s- 1 3 . 29 10 15 s- 1 = 0 . 0625 n 2 2 = 16 n 2 = 4 005 10.0 points The n and quantum numbers of the last electron of an element are n = 4 and = 2 . The element is 1. an f-transition metal. 2. a noble gas....
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This note was uploaded on 08/30/2010 for the course CH 52385 taught by Professor Vandenbout during the Spring '09 term at University of Texas at Austin.

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McCord - 2010 - Exam 1-solutions - Version 238 Exam 1...

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