Using the greens function of the diffusion equation p

Using the Green’s function of the diffusion equation,
P ( r , t | r
0
, 0 ) =
1
( 4 π Dt )
3 / 2
e
-| r - r
0
|
2
/ 4 Dt
analyze the time dependence of the electron density distribution.
Is it correct to use this Green’s function for the case of the electron creation at the surface?

43
Solution 4.3: For the density distribution at time t one has (by definition)
n ( r , t ) =
Z
d
3
r
1
P ( r , t | r
1
, 0 ) n ( r
1
, 0 ) .
The solution is substantially simplified because P ( r , t | r
0
, 0 ) depends only on the difference ( r -
r
1
) . Then, in general, one can use the Fourier method.
However, this problem is even simpler since this function depends only on
e
-| r - r
1
| / 4 Dt
= e
- ( x - x
1
)
2
/ 4 Dt
· e
- ( y - y
1
)
2
/ 4 Dt
· e
- ( z - z
1
)
2
/ 4 Dt
,
while
n ( r
0
, 0 ) ∝ e
- x
2
1
/ 2 σ
· e
- y
2
1
/ 2 σ
· e
- z
2
1
/ 2 σ
.
Thus we have to analyze only one integral,
I ( x ) ≡
1
2 π
√
2 σ Dt
Z
∞
- ∞
dx
1
e
- ( x - x
1
)
2
/ 4 Dt
e
- x
2
1
/ 2 σ
=
1
p
2 π ( σ + 2 Dt )
e
- x
2
/ 2 ( σ + 2 Dt )
.
As a result,
n ( r , t ) =
N
[ 2 π ( σ + 2 Dt )] 3 / 2
exp
•
-
| r |
2
2 ( σ + 2 Dt )
‚
.
This expression cannot be used for the case of surface excitation since the Green’s function for
that problem has different form.
Problem 4.4: In the presence of electrical field, the electrons acquire a constant (drift) velocity,
proportional to the applied electric field, v
d
= μ E . The quantity μ is called the mobility .
Let N electrons be injected at the plane x = 0, t = 0, and then diffuse in the presence of the
electrical field creating the drift velocity v
d
k x .
Find the distribution of electrons at time t .
Solution 4.4: The particle current in the presence of the drift is
j = n ( x ) v
d
- D
dn
dx
.
Thus the diffusion equation acquires the form:
dn
dt
+
d j
dx
= 0 →
dn
dt
+ v
d
dn
dx
- D
d
2
n
dx 2
= 0 .
It is natural to introduce a new variable, ξ = x - v
d
t and search solution as n ( ξ , t ) . Since
dn ( ξ , t )
dt
=
∂ n ( ξ , t )
∂ t
- v
d
∂ n ( ξ , t )
∂ξ
.

44 CHAPTER 4. DEFECTS AND DIFFUSION
On the other hand,
v
d
dn
dx
= v
d
∂ n ( ξ , t )
∂ξ
∂ξ
∂ x
= v
d
∂ n ( ξ , t )
∂ξ
.
In this way we return to the well known diffusion equation,
∂ n
∂ t
= D
∂
2
n
∂ξ
2
having the solution
n ( x , t ) =
n
0
√
4 π Dt
exp -
( x - v
d
t )
2
4 Dt
¶
.
Here n
0
≡ N / A is the initial surface density of injected particles.
Problem 4.5: Frenkel defects: Show that the number n of interstitial atoms in equilibrium
with n lattice vacancies in a crystal having N lattice points and N
0
possible interstitial positions
is given by the equation
E
I
= k
B
T ln
£
( N - n )( N
0
- Nn ) / n
2
/
,
whence for n ¿ N , N
0
, we have n ’ ( NN
0
)
1 / 2
exp ( - E
I
/ 2 k
B
T ) . Here E
I
is the energy necessary
to remove an atom from a lattice site to an interstitial position.
Solution 4.5: The internal energy of formation of n Frenkel defects is U = nE
I
. The number
of ways to pick n from N is
N !
n ! ( N - n ) !
.
The number of ways to put n atoms into N
0
sites is
N
0
!
n ! ( N 0 - n ) !
.
Hence, the entropy is
S = k
B
ln
N !
n ! ( N - n ) !
+ ln
N
0
!
n ! ( N
0
- n ) !
¶
.
Using the Stirling’s formula we get
ln
N !