Atomic/Molecular
Collision
Theory
Supplementary
Notes
for
EP271
M.P.
Bradley
(Please
note
that
these
notes
are
based
on
earlier
handwritten
notes
by
Prof.
Adam
Bourassa,
with
some
additions/extensions)
Atomic/Molecular
Collisions
&
the
Approach
to
Equilibrium
Kinetic
theory
gives
us
some
insight
into
the
equilibrium
properties
of
gases.
We
have
already
seen
how
atomic
or
molecular
collisions
with
the
chamber
walls
give
rise
to
the
measurable
gas
pressure.
More
deeply,
it
is
the
collisions
between
atoms
and
molecules
which
lead
to
the
development
of
a
stable,
equilibrium
distribution
(such
as
the
Maxwell‐Boltzmann
distribution).
In
this
set
of
supplementary
notes
we
consider
some
basic
notions
of
collision
theory.
Collision
Cross
Section
and
Collision
Probability
Imagine
a
column
of
gas
of
total
cross‐sectional
area
A
and
length
Δx
filled
with
gas
atoms
or
molecules
of
individual
cross‐section
σ
(measured
in
units
of
m2,
and
typically
very
small
σ~10‐20
m2).
Then
as
viewed
by
an
“incident”,
or
“projectile”
particle
traversing
the
column,
the
effective
projected
area
covered
by
“target”
atoms
is
Covered
area:
Acov = σn(AΔx) Where
n
is
the
“number
density”
of
atoms
in
the
gas
(i.e.
number
of
atoms
per
m3)
If
the
gas
has
a
low
enough
density
that
the
“target”
atoms
don’t
“occlude”
or
block
€
on
another,
the
probability
that
an
incident
projectile
atom
will
strike
a
target
atom
while
travelling
through
the
column
is
simply
give
by
the
ratio
of
the
area
“occupied”
by
target
atoms
to
the
total
cross‐sectional
area
of
the
gas
column,
i.e.
Probability
of
Collision:
P =A cov / A = σnΔx Now
if
N
is
the
total
number
of
incident
projectile
atoms
incident
on
the
column
of
target
gas,
then
the
number
of
atoms
“knocked
out”
of
the
incident
beam
is
simply
€
given
by
the
product
of
the
number
of
incident
atoms
multiplied
by
the
probability
of
collision,
i.e. ΔN = −NP = −N(σnΔx) .
We
can
then
re‐arrange
this
to
get
the
first‐
order
differential
equation:
dN /dx = −σn The
solution
to
this
simple
differential
equation
is
N(x) = N 0e−σnx €
€
We
can
elaborate
on
this
idea
a
bit.
If
we
imagine
that
we
have
a
gas
composed
of
hard‐sphere
atoms
(like
billiard
balls),
then
the
collision
cross
section
would
be
€
2
σ = π ( rincident + rtarget ) = πD 2 where
D
=
rincident/2
=
rtarget/2
is
the
effective
atomic
or
€
molecular
“diameter”
for
collisions.
Since
atoms
and
small
molecules
have
characteristic
sizes
~
a
few
Å,
we
expect
effective
collision
cross
sections
σ
~
10‐20
m2.
Now,
it
is
useful
to
know
(and
to
be
able
to
calculate
or
estimate)
the
average
distance
which
a
gas
atom
or
molecule
travels
between
collisions.
To
this
end,
we
can
define
a
“mean‐free‐path”
λ
by:
total distance travelled in time t
λ=
= mean distance between collisions number of collisions in time t
Now
for
an
atom/molecule
with
average/mean
speed
v = v the
total
distance
travelled
in
time
t
is v t Following
the
arguments
we
used
previously,
the
number
of
collisions
occurring
€
during
this
time
is
(πD 2v t)n €
vt
1
1
So
from
this
we
get
λ =
.
So
we
see
that
with
this
we
can
re‐
=
=
2
2
πD v tn (πD ) n σn
€
write
our
equation
from
above
in
terms
of
the
mean
free
path
λ:
N(x) = N 0e−x / λ €
Now
we
have
to
slightly
adjust/correct
our
formula
for
the
mean
free
path
λ.
We
assumed
that
we
had
stationary
targets,
but
in
fact
the
“target”
atoms
are
other
gas
atoms
which
are
also
moving,
with
mean
velocity
v =< v > .
So
we
have
to
account
for
this.
How
?
Well,
in
a
collision
between
a
moving
projectile
and
a
moving
target,
what
counts
is
the
relative
velocity.
The
magnitude
of
the
vector
relative
velocity
is
v rel = v12 + v 22 ,
so
the
average
relative
velocity
for
a
collision
between
two
atoms
€
€
€
€
€
from
the
same
gas
is
v rel = v 2 + v 2 = 2v .
So
we
need
to
multiply
the
mean
velocity
term
which
appears
in
our
expression
for
mean
free
path
by
a
factor
of
2 ,
1
So
we
get
finally
λ =
2σn
€
€
N
P
Now
we
can
use
the
ideal
gas
law
here
since
we
know
that
n = =
,
which
gives
V kB T
€
k T
us
λ = B 2σP
€
Let
us
do
a
calculation
for
oxygen
molecules
O2
at
T=
300
K
and
sea
level
(P=1
atm).
(We
can
assume
D=3
Å):
(1.38 ×10 J /K )(300K) ≅ 10
2π ( 3 ×10 ) (101325Pa)
−23
We
get
λ =
−7
−10 2
o
m = 1000 Α = 0.1µm .
While
this
might
seem
like
a
small
distance
it
is
important
to
note
that
it
is
much,
much
larger
than
the
size
of
the
individual
atoms
or
molecules
in
the
gas.
λ
€ We
also
compute
the
mean
time
between
collisions
using
t = .
For
our
example
of
v
8 kB T
O2
(m=32
u)
at
T=300
K,
we
have
v = v =
.
Using
the
molecular
mass
π m
M
m=
= 32 ×10−3 kg /N A = 5.3 ×10−26 kg /molecule€
in
the
above
formula,
we
get
NA
€
€
€
€
−23
8 (1.38 ×10 J€/K )( 300K )
v= v =
= 446m /s .
π
5.3 ×10−26 kg
This
gives
us
a
mean
time
between
collisions
of
λ 0.1µm
t avg = =
= 2.2 ×10−10 s = 0.22ns .
So
the
average
time
between
collisions
is
v 446m /s
very,
very
short,
and
average
molecule
undergoes
collisions
with
a
frequency
1
f collision = = 4 ×10 9 s−1 ,
i.e.
4
billion
collisions
per
second!
t
It
is
also
interesting
to
look
at
the
approximate
mean
free
paths
of
gas
molecules
(O2
or
N2)
in
the
atmosphere,
as
function
of
the
altitude
(table
from
calculations
of
Prof.
Adam
Bourassa)
Altitude
MFP
Sea
level
(P=1
atm)
λ=
0.1
µm
100
km
λ=
16
cm
300
km
λ=
20
km