This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Exam 1 mccord (51600) 1 This printout should have 34 questions. Multiplechoice 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 Hatom`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....
View
Full
Document
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.
 Spring '07
 Fakhreddine/Lyon
 Chemistry

Click to edit the document details