25.
(a) Promoting one of the electrons (described in problem 23) to a notfully occupied higher level, we
Fnd that the conFguration with the least total energy greater than that of the ground state has the
n
= 1 and 2 levels still Flled, but now has only one electron in the
n
= 3 level; the remaining two
electrons are in the
n
= 4 level. Thus,
E
frst excited
=2
E
1
+2
E
2
+
E
3
E
4
µ
h
2
8
mL
2
¶
(1)
2
µ
h
2
8
mL
2
¶
(2)
2
+
µ
h
2
8
mL
2
¶
(3)
2
µ
h
2
8
mL
2
¶
(4)
2
= (2+8+9+32)
µ
h
2
8
mL
2
¶
= 51
µ
h
2
8
mL
2
¶
.
(b) Now, the conFguration which provides the next higher total energy, above that found in part (a),
has the bottom three levels Flled (just as in the ground state conFguration) and has the seventh
electron occupying the
n
= 5 level:
E
second excited
E
1
E
2
E
3
+
E
5
µ
h
2
8
mL
2
¶
(1)
2
µ
h
2
8
mL
2
¶
(2)
2
µ
h
2
8
mL
2
¶
(3)
2
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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.
 Fall '08
 SPRUNGER
 Physics, Energy

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