This preview shows page 1. Sign up to view the full content.
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 = = ...
View Full Document
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
- Fall '08