p41_026 - 26. (a) Using Eq. 40-20 (see also problem 27 in...

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Unformatted text preview: 26. (a) Using Eq. 40-20 (see also problem 27 in Chapter 40) we find that the lowest five levels of the rectangular corral (with this specific “aspect ratio”) have energies E1,1 = 1.25, E1,2 = 2.00, E1,3 = 3.25, E2,1 = 4.25, and E2,2 = 5.00 (all of these understood to be in “units” of h2 /8mL2 ). It should be noted that the energy level we denote E2,2 actually corresponds to two energy levels (E2,2 and E1,4 ; they are degenerate), but that will not affect our calculations in this problem. The configuration which provides the lowest system energy higher than that of the ground state has the first three levels filled, the fourth one empty, and the fifth one half-filled: Efirst excited = 2E1,1 + 2E1,2 + 2E1,3 + E2,2 = 2(1.25) + 2(2.00) + 2(3.25) + 5.00 which means (putting the “unit” factor back in) the energy of the first excited state is Efirst 18.00(h2 /8mL2 ). excited = (b) The configuration which provides the next higher system energy has the first two levels filled, the third one half-filled, and the fourth one filled: Esecond excited = 2E1,1 + 2E1,2 + E1,3 + 2E2,1 = 2(1.25) + 2(2.00) + 3.25 + 2(4.25) which means (putting the “unit” factor back in) the energy of the second excited state is Esecond 18.25(h2 /8mL2 ). excited (c) Now, the configuration which provides the next higher system energy has the first two levels filled, with the next three levels half-filled: Ethird excited = 2E1,1 + 2E1,2 + E1,3 + E2,1 + E2,2 = 2(1.25) + 2(2.00) + 3.25 + 4.25 + 5.00 which means (putting the “unit” factor back in) the energy of the third excited state is Ethird 19.00(h2 /8mL2 ). (d) The energy states of this problem and problem 24 are suggested in the sketch below: third excited 19.00(h2 /8mL2 ) second excited 18.25(h2 /8mL2 ) first excited 18.00(h2 /8mL2 ) ground state 17.25(h2 /8mL2 ) excited = = ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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