Page 276 - Essentials of physical chemistry

238 Essentials of Physical Chemistry

probability events. Therefore if we integrate our probability function over all possible values of the
ð
npx

variable, the result should be set to 1. Thus we have 1 ¼ c*cdx. Here c* ¼ iB sin and
L
L
ð
2 2 2 npx
( iB)(iB) ¼þB , so the ‘‘*’’ makes the result a real number. B sin dx ¼ 1. This condi-
L
0
tion will allow us to evaluate B! It is common practice to neglect the ‘‘i’’ if it is a direct factor since
we have seen that it will disappear whenever an integral for a probability is carried out but it has
served here to illustrate the need for the complex conjugate. Thus we have to do the integral and
L
ð L npx ð 1 cos 2npx
2
solve for the B coefficient. B sin 2 dx ¼ 1 ¼ B 2 L dx using the half-angle
0 L 0 2
formula from trigonometry. We will have to split this into two integrals.
8 3L 9

2npx >
2
8
ð L ð L > sin >
>
9
>
>
2
1 < 2npx = B 2 < 6 L 7 = B 2 B L
2
1 ¼ B dx cos dx ¼ L 6 7 ¼ [L 0] ¼
L 4 2np 5 2 2
2 : ; 2 > >
> >
0 0 > >
L
:
;
0
r ffiffiffi
2
B L 2
If ¼ 1, then we have found B ¼ so in conclusion we have the full solution (neglecting i)
2 L
!
r ffiffiffi
2 2
2 npx n h
sin and , n ¼ 1, 2, 3, 4, 5, ..., n
n
L L 8mL
c (x) ¼ E n ¼ 2
Note also that the n ¼ 0 case is the ‘‘no-particle’’ case and the energy is quadratic in n for n > 0.
Let us consider the ultraviolet spectrum of butadiene (Table 11.1). Assuming two electrons
loosely spin-paired in each orbital and neglecting the interaction between the electrons, we can fill
two energy levels with four electrons. Then the lowest energy excitation will be from the highest
occupied molecular orbital (HOMO) n ¼ 2 to the lowest unoccupied molecular orbital (LUMO)
n ¼ 3. We can estimate the wavelength of this transition as a one-electron jump from n ¼ 2
c (9 4)h 8mL c
2 2
to n ¼ 3. DE ¼ E 3 E 2 ¼ hn ¼ h ¼ ) l ¼ , but what shall we use for L?
l 8mL 2 5h
We can estimate L as three times the length of the aromatic bond in benzene since the middle
bond is part of the pi-electron system as 3 (1.4 Å) ¼ 4.2 Å. Then we can find l
2
8
2
8mL c 8(9:10938215 10 31 kg)(4:2 10 10 m) (2:99792458 10 m=s)
as l ¼ ¼ and so
5h 5(6:62606896 10 34 J s)
l ffi 1:1632 10 7 m ¼ 1:1632 10 5 cm ¼ 1163:2 10 8 cm ¼ 1163:2 Å. This is too far
into the vacuum ultraviolet so let us treat L as a variable parameter and find what value of L will
fit the experimental wavelength of 2170 Å [2, p. 103].
TABLE 11.1
Absorption Bands of Selected trans-Polyenes
Compound # p Electrons l (Å)
Ethylene 2 1625
Butadiene 4 2170
Hexatriene 6 2510
Octatetraene 8 3040
Source: Davis, J.C., Advanced Physical Chemistry, The
Ronald Press Co., New York, 1965, p. 103.