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Unformatted text preview: Chem 113A HW4 Solutions 1. i. E = h 2 n 2 8 ma 2 E = h 2 8 ma 2 ( n 2 2 n 2 1 ) = (6 . 626 * 10 34 Js ) 2 8 * (9 . 11 * 10 31 kg ) * (1 * 10 9 m ) 2 (4 1) = 1 . 807 10 19 J ii E = h 2 8 ma 2 ( n 2 2 n 2 1 ) = (6 . 626 * 10 34 Js ) 2 8 * (65 kg ) * (1 * 10 9 m ) 2 (4 1) = 2 . 533 10 51 J This result shows a very small energy difference between energy levels. It is 1.583*10 32 eV. This means that your quantum state can not be observed in the box of this size. Clearly, we are classical beings in these conditions. 2. The energy for a particle in a 2D box with equal sides (a=b) is given by: E = h 2 8 m ( n 2 x a 2 + n 2 y b 2 ) = h 2 8 ma 2 ( n 2 x + n 2 y ) i. The ground state can only be defined by one combination of numbers, and so the system is singly degenerate with n x = 1 and n y = 1. ii. For the first excited state, there are two combinations that would give the same energy value, so this state is doubly degenerate with n x = 1 and n y = 2 or n x = 2 and n y = 1....
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This note was uploaded on 12/05/2011 for the course CHEM 113A taught by Professor Lin during the Fall '07 term at UCLA.
 Fall '07
 Lin
 Quantum Chemistry

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