Chemistry Reference
In-Depth Information
EXAMPLE 6-1
Since the probability of having either heads or tails for a fair coin is equal, it is not unreason-
able to expect that the mean displacement of a random walker after a large number of steps
would be zero. Show that
Eq. (6-19)
yields this expectation.
Solution
Given that the length of each step is equal to L, we have:
Average displacement 5 hnLi
ð
N
ð
N
nL
e
2n
2
=2N
nLP(n)dn 5
p
2πN
dn
2
N
2
N
ð
N
h
i
N
2
N
50
5
2
2NL
2n
N
e
2n
2
=2N
dn 5
2
2NL
e
2n
2
=2N
p
2πN
p
2πN
2
N
What is missing in the above analysis is the time scale associated with each
step. Now let's define
as the time that each step takes. After the random walker
has walked over a period of time t, the total number of steps, N, is given by
τ
t
τ
N
5
(6-20)
By applying
Eq. (6-20)
to
Eq. (6-19)
(see Problem 6-2), one obtains the fol-
lowing equation for the mean square displacement of the random walk after N
steps:
t
ð
N
L
2
2
2
n
2
L
2
P
hð
nL
Þ
i 5
ð
n
Þ
dn
5
2
(6-21)
τ
2
N
is essentially the one-dimensional self-diffusion coefficient of the
random walker. In the following example, Fick's laws will be used to show such
correlation.
L
2
2
The term
τ
EXAMPLE 6-2
Consider a one-dimensional diffusion process involving a point source of particles located at
the origin of the x-axis (
Figure 6.4
). Show that the mean square displacement of the particles
over a period of time t is 2Dt where D is the self-diffusion coefficient of the particles.
Solution
Assuming that the mutual diffusion coefficient D defined in Fick's first law is both time and
concentration independent and that the particle concentration is very low (i.e., no interac-
tions between the particles), D in this case is the self-diffusion coefficient of particles that
