Environmental Engineering Reference
In-Depth Information
For simplicity, we assume all particles to be spherical and to have the same radius
r
0
. This radius is taken to be sufficiently large such that
e
2
T
r
0
.
(1.63)
This criterion allows us to consider a particle as bulk matter, rather than needing to
describe its microscopic properties. The addition of a single electron to the aerosol
particle makes only a slight difference to the electric potential of the particle. We
use this property to write the relationship between the number densities of parti-
cles
n
Z
and
n
Z
C
1
that possess charges
Z
and
Z
C
1, respectively. By analogy with
the Saha distribution (1.52), we have
2
m
e
T
2
exp
,
3/2
n
Z
N
e
n
Z
C
1
W
Z
T
D
(1.64)
π
„
2
where
W
Z
is the electron binding energy for the particle with charge
Z
,
N
e
is the
electron number density, and the factor 2 accounts for the electron statistical weight
(two spin projections). The electron binding energy of a charged particle
W
Z
is the
sum of the electron binding energy for the neutral particle
W
0
of a given material
and the potential energy of the charged particle. Using the electric potential for a
particle of charge
Z
C
1/2 (the average between
Z
and
Z
C
1), we have
Z
e
2
r
0
1
2
W
Z
D
W
0
C
C
.
Substituting this into (1.64) transforms it to the form
exp
!
.
Z
2
e
2
r
0
T
2
m
e
T
2
3/2
1
C
n
Z
N
e
n
Z
C
1
W
0
T
D
(1.65)
π
„
2
This relation gives the charge distribution of the particles. If the average charge
is large, this distribution is sharp. Specifically, introducing
n
0
, the number density
of neutral particles, (1.65) leads to
n
Z
1
A
exp
n
0
A
Z
exp
,
Ze
2
r
0
T
Z
2
e
2
2
r
0
T
n
Z
D
D
2
)]
3/2
exp(
where
A
W
0
/
T
). For charges that are close to the
average, this relationship is conveniently written in the form
D
(2/
N
e
)
[
m
e
T
/(2
π
„
n
Z
exp
!
,
(
Z
Z
)
2
n
Z
D
2
Δ
Z
2
Z
2
r
0
T
/
e
2
where
1
because of (1.63). The average charge of the particles
follows from the relation
Ze
2
/(
r
0
T
)
Δ
D
D
ln
A
,whichgives
ln
"
2
N
e
#
.
m
e
T
2
3/2
exp
r
0
T
e
2
W
0
T
Z
D
(1.66)
π
„
2
