Chemistry 1410, Hour Exam 1, Solution Key
1) a) Appeal to the 1D Particle in the Box (PinB) model for the single electron energy levels of the pi
electrons in hexatriene. Furthermore: respect the Pauli Exclusion Principle, putting no more than two
(spin-paired) electrons in each single-electron spatial state. This leads to the diagram shown in Fig. A1,
and thus the conclusion that the highest energy pi electron goes into the PinB level n=3 when the system
is in its electronic ground state configurations (electrons are packed into the lowest single-electron states
possible without violating the Pauli Exclusion Principle).
b) The lowest energy transition possible would be for an electron occupying n=3 to make a transition into
the (previously unoccupied) level n=4. Denoting En as the PinB energy eigenvalue corresponding to
level n, then 'E43
E4 E3 is the increase in electronic energy incurred when an electron makes the
transition from n 3 o n 4 . This energy has to supplied by the absorbed photon, i.e., E ph 'E43 .
h2
, where me is the electron mass and L is
8me L2
Recall that for an electron moving in a 1D PinB, En
the effective box length. Using the same arguments used for butadiene (cf. Fig. 1 on the exam) and
octatetraene (see class notes), we arrive at the approximation for all-trans hexatriene (with 6 carbon
atoms) that L # 6 R , with R being the effective average C-C bond length: cf. Fig. A2. Now we can
calculate:
E ph
'E43
Thus: D
c)
h2
[42 32 ]
2
8me (6 R)
7 h2
36 8me R 2
7 / 36 # 0.194 .
Since
h2
=1.6x105cm-1,
2
8me R
7
˜ 1.6 u 105 cm 1 3.1 u 104 cm 1 .
36
[3.1 u 104 ]1 cm 3.23 u 105 cm 3, 233 Å.
E ph
then
the
This
absorbed
corresponds
photon
to
a
energy
wavelength
is
of
1
Fig. A1. Ground electronic state configuration for all-trans hexatriene,
assuming a 1D Particle in a Box model for the single electron states.
Fig. A2. Mapping the all-trans hexatriene molecule to a 1D Particle in a Box:
determining the box length L=5R + 2(R/2) = 6R.
2
3) a) For a generic 1D harmonic oscillator characterized by mass m and force constant k, the allowed
1
(n )=Z , n=0,1,2,..., with Z { k / m . The energy of the 2nd
2
5
excited state thus corresponds to n=2, or E2
=Z .
2
energy eigenvalues are given by En
b) From the class notes (or Engel textbook), the energy eigenfunction corresponding to E2 is
1/ 4
\ 2 ( x)
1 § mZ ·
¨
¸
2 2 © S= ¹
§ mZ ·
­ mZ 2 ½
H 2 ¨¨
x ¸¸ exp ®
x ¾
¯ 2= ¿
© = ¹
with Hermite polynomial H 2 ( y ) { 4 y 2 2 .
A sketch is provided in Fig. A3. [Note the following properties: i) \ 2 ( x) is even about x=0; ii)
\ 2 ( x) o 0 as x o rf ; ii) \ 2 ( x) is “balanced” about the value \ 2
0 , in that
³
f
\ 2 ( x)dx
f
0
(why?).]
Fig. A3: sketch of \ 2 ( x) .
c)
i) Let a0
0.775, a1
³
f
f
0.5, a2
dx\ * ( x)\ ( x)
0.387 . Then:
³
f
*
f
³
dx a0M0 a1M1 a2M2 a0M0 a1M1 a2M 2 f
f
dx a0M0 a1M1 a2M2 a0M0 a1M1 a2M 2 , [A1]
3
where the second line follows from the first because the coefficients a j and the energy eigenfunctions
M j ( x) are real-valued.
Note next that
³
eigenfunctions are orthogonal, and that
f
f
³
f
f
dxM j ( x)Mk ( x)
0 for j z k , because the energy
dxMk ( x)Mk ( x) 1 for k
0,1, 2 , since these eigenfunctions
are stated to be “normalized”. Thus, all the integrals in Eq. (A1) can easily be performed, with the result
that:
³
f
f
dx\ * ( x)\ ( x)
2
¦a
2
j
.7752 .52 .387 2
1
j 0
QED.
ii) The probability Pn that a measurement of energy yields En is given by an2 .
P1
0.52
iii) E !
In particular,
0.25 .
2
¦a
n
2
En
1.05=Z .
n 0
4