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hw4_ans - Chem 113A HW4 Solutions 1 i E = h2 n 2 8ma2 ii E...

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Chem 113A HW4 Solutions 1. i. E = h 2 n 2 8 ma 2 Δ 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 Δ 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 2-D 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. 3. For a particle in a 1-D box of length a, the wavefunction is given by: ψ ( x ) = ( 2 a ) 1 / 2 sin nπx a
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