MT 2 Sol - Midterm 2 Solutions 4/16/2010 1. (a) Answer: ii,...

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Midterm 2 Solutions 4/16/2010 1. (a) Answer: ii, iii, iv and v See problem set 5 question 3 and PS 6 question 3 Scoring: 5 x - 5 y , where x are the number of correct answers and y are the number of incorrect answers. Note that negative scores were treated as 0. (b) Answer: ii, iii, and v See PS 6 problem 4 and PS 7 problem 5 Scoring: 5 x - 5 y (c) Answer: i, iii, and v We are looking at a neutral 2-electron atom. Because the average electron-nuclear separation for the electron in the 4p orbital is so much greater than that of the electron in the 1s orbital we can consider the 1s electron to perfectly shield the nucleus. This gives E 0 / 16 as a crude estimate of the IP. Remember that the eigenvalues of ˆ L 2 are ~ 2 l ( l + 1). Furthermore the l quantum number for the 1s state is 0 while for the 4p state it is 1. We have given no information on how this state was prepared therefore it could be either a singlet or a triplet. Scoring:
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This note was uploaded on 05/14/2010 for the course CHEM 120ACHEM taught by Professor Chandler during the Spring '10 term at University of California, Berkeley.

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MT 2 Sol - Midterm 2 Solutions 4/16/2010 1. (a) Answer: ii,...

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