Exam I Solutions - MIT OpenCourseWare http/ocw.mit.edu...

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MIT OpenCourseWare http://ocw.mit.edu 5.111 Principles of Chemical Science Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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First Hour Exam 5.111 5.111 Write your name below . Do not open the exam until the start of the exam is announced. The exam is closed notes and closed book. 1. Read each part of each problem carefully and thoroughly. 2. Read all parts of each problem. MANY OF THE LATTER PARTS OF A PROBLEM CAN BE SOLVED WITHOUT HAVING SOLVED EARLIER PARTS. However, if you need a numerical result that you were not successful in obtaining for the computation of a latter part, make a physically reasonable approximation for that quantity (and indicate it as such) and use it to solve the latter parts. 3. A problem that requests you to “calculate” implies that several calculation steps may be necessary for the problem’s solution. You must show these steps clearly and indicate all values, including physical constants used to obtain your quantitative result. Significant figures must be correct. 4. If you don’t understand what the problem is requesting, raise your hand and a proctor will come to your desk. 5. Physical constants, formulas and a periodic table are given on the last page. You may detach this page once the exam has started . Suggested time 1. 12 minutes (22 points) _____________ 2 . 10 minutes (16 points) ______________ 3 . 19 minutes (38 points) ______________ 4 . 9 minutes (24 points) ______________ Total (100 points) _________________ Name ___________________________________
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Page 2 of 10 pages 1. (22 points) The photoelectric effect A beam of light with an intensity of 15 W is incident on a copper plate ( φ = 7.43 x 10 -19 J). Electrons with a minimum wavelength of 3.75 x 10 -10 m are ejected from the surface of the copper. ( a) (12 points) Calculate the frequency of the incident light. e - E i λ = 3.75 x 10 -10 m φ = 7.43 x 10 -19 J K.E. of electron: k g m 2 s -2 λ = h p = h = 6.626 x 10 -34 Js = 1.76 7 x 10 -24 kgms -1 p λ 3.75 x 10 -10 m E = p 2 = (1.76 7 x 10 -24 kgms -1 ) 2 = 1.71 38 x 10 -18 J 2m e 2 (9.109 x 10 -31 kg) E i = φ + KE = 0.743 x 10 -18 + 1.71 4 x 10 -18 J = 2.45 7 x 10 -18 J E = h ν ν = 2.45 7 x 10 -18 J = 3.70 8 x 10 15 s -1 ν
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This note was uploaded on 09/25/2010 for the course BIOLOGY Biology 10 taught by Professor D'arville during the Spring '10 term at Rutgers.

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Exam I Solutions - MIT OpenCourseWare http/ocw.mit.edu...

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