ΔE =
hc
λ
=
6:626 Â 10
−34
 Js
À
Á
2:998 Â 10
8
 ms
−1
À
Á
250 Â 10
−9
 m
À
Á
= 7:946 Â 10
−19  J
Assuming that the 1 ! 2 transition is associated with the absorption wavelength of 250 nm, we rearrange Equation 5.34 to determine the moment of inertia.
I =
h
2 2
2
− 1
2
À
Á
ΔE 8π 2
À
Á =
6:626 Â 10
−34
 Js
À
Á 2 3
ð Þ
7:946 Â 10
−19
 J
À
Á
8π
2
À
Á = 2:102 Â 10
−50
 Js
2
= 2:102 Â 10
−50
 kgm
2
(Note: 1 Js
2 = 1 kgm
2
)
Finally, the radius can be estimated using Equation 5.35.
r =
ffiffiffiffiffiffiffi
I
m e
s
=
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
2:102 Â 10
−50
 kgm
2
9:109 Â 10
−31
 kg
s
= 1:519 Â 10
−10  m = 0:152  nm
This is close to the experimentally measured value of around
0.25 nm.
C 6 H 6
C
C
C
C
C
C
Unhybridized
p-orbital
containing
one electron
(a)
CC σ-framework
(b)
E = 4h 2
E = h 2
8π 2 I
8π 2 I
m = ±2
m = ±1
m = 0
E = 0
Figure 5.17 (a) Overlap of unhybridized p-orbitals on each carbon atom produces
a π molecular orbital resulting in the delocalization of electron density along the
benzene ring. (b) The particle-in-a-box model as applied to the benzene molecule
results in an energy level diagram showing electron pairs in three levels. The two
energy levels corresponding to m = ±1 are doubly degenerate.
SIMPLE MODELS DESCRIBING ELECTRONIC STRUCTURE 169