ch301 exam 1 - Exam 1 mccord (51600) 1 This print-out...

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Unformatted text preview: Exam 1 mccord (51600) 1 This print-out should have 34 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. c = 3 . 00 10 8 m/s h = 6 . 626 10 34 J s m e = 9 . 11 10 31 kg R = 3 . 29 10 15 s 1 R = 2 . 178 10 18 J N A = 6 . 022 10 23 mol 1 g = 9 . 81 m/s 2 E = h c = / = h p = h mv 1 2 mv 2 = h- E n =- R n 2 (hydrogen atom) = R parenleftbigg 1 n 2 1- 1 n 2 2 parenrightbigg- h 2 2 m d 2 d x 2 + V ( x ) = E 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 What is the frequency of light with a wave- length of 4.0 10 7 m? 1. 120 s 1 2. 3 . 10 14 s 1 3. 1 . 3 10 15 s 1 4. 7 . 5 10 14 s 1 5. 3 . 10 14 s 1 Explanation: 002 10.0 points Assume n 1 and n 2 are two adjacent energy levels of an atom. The emission of radiation with the longest wavelength would occur for which two values of n 1 and n 2 ? 1. 5,4 2. 7,6 3. 8,7 4. 3,2 5. 2,1 6. 6,5 7. 4,3 Explanation: The frequency of a photon emitted when an electron moves between levels n 1 and n 2 is given by the Rydberg equation: = R parenleftbigg 1 n 2 1- 1 n 2 2 parenrightbigg , where R = 3 . 29 10 15 Hz. The emission of radiation with the longest wavelength corre- sponds to that with the smallest frequency. From inspection of the formula above we see that is smallest when n 1 = 8 and n 2 = 7. Conceptual Solution : E = h = h R parenleftbigg 1 n 2 1- 1 n 2 2 parenrightbigg gives the energy of the pho- tons emitted. The emission of radiation with the longest wavelength corresponds to pho- tons with the smallest energy. From the Bohr frequency condition the energy of the emit- ted photon must be equal to the difference in energy between the higher and lower lev- els. An energy level diagram for the H-atom-`i`i`iiv 6v**,3-` /iiVi]\ ViVV Exam 1 mccord (51600) 2 shows that as the energy levels get higher, the gaps between them converge; of the transi- tions listed, the two adjacent levels which are the closest together are n 1 = 8 and n 2 = 7. Thus a transition from n 1 = 8 to n 2 = 7 will result in the emission of a photon with the smallest energy, hence the longest wave- length. 003 0.0 points If more points are awarded on this assign- ment, would you like them added to your score? 1. NO, leave my score alone, I prefer the lower score 2. YES, I would like the points and the higher score....
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This note was uploaded on 08/17/2011 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas at Austin.

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ch301 exam 1 - Exam 1 mccord (51600) 1 This print-out...

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