Luiz
Leal
Oak
Ridge
National
Laboratory
Lectures
Presented
at
the
Nuclear
Engineering
Department
of
the
Massachusetts
Institute
of
Technology
(MIT)
Courtesy of Luiz Leal, Oak Ridge National Laboratory. Used with permission.
Neutron
Interaction
Time
Independent
Transport
Equation
for
 ˆ
Φ(E , r ,Ω
)

dΩ̂
'
dE'
Σ
(E'→
E,
Ω̂
'
→
Ω)Φ(E',r
',Ω̂) + S
∫ ∫
s
∞
Ω̂.∇Φ + Σ t Φ =
€
€
€
€
4π
0
Σ t Macroscopic
total
cross
section
Σ s Macroscopic
scattering
cross
section
How
are
these
quantities
treated?
(a) Measurement
by
time‐of‐flight
machine,
reactors,
etc;
(b) Evaluated
using
nuclear
physics
models;
(c)
Generated
using
nuclear
physics
codes
(Poor
Choice!)
How
is
nuclear
physics
applied
to
get
the
cross
sections?
Neutron
Interaction
n
+
target
→ ???
In
all
three
regions
there
are
always:
SCATTERING
and
REACTION
Region
1:
High
energy
neutrons
–
direct
and
compound
nucleus
formation;
Region
2:
Resonance
region
(resolved
and
unresolved)
–
Definitely
compound
nucleus
formation
and
some
direct
interactions;
Region
3:
Chemical
region
–
neutron
energy
of
incident
neutron
is
comparable
to
the
chemical
binding
energy
of
atoms
in
the
molecules;
The
neutron
reaction
cross
sections
such
as
(n,
fission),
(n,
gamma)
are
affected
only
by
the
motion
of
the
target
atoms
(Doppler
broadening
effects).
In
the
treatment
of
the
neutron
scattering
cross
section
in
the
chemical
region
one
must
consider:
(a) Kinetic
energy
of
the
incident
neutrons
is
comparable
to
the
binding
energy
of
the
atom
in
the
target
(molecule,
solid,
liquid).
ATOMS
ARE
NOT
FREE!!!
(b) Neutron
wavelength
is
of
the
order
of
the
interatomic
spacing
in
molecules
of
crystals;
(c) Scattering
from
various
nuclei
in
the
same
molecule
of
crystal
interfere
(coherent,
incoherent);
(d) Neutron
may
GAIN
or
lose
energy;
€
Born
approximation
And
Fermi
pseudo
potential
Consider
the
situation:
Scattering
by
a
single
spinless
nucleus

ψ (r ) at
the
detector
satisfy
the
equation:
 2 2
 

− ∇ + V (r ) ψ (r )=Eψ (r )
 2µ

€
Or


[ ∇ + k ] ψ (r )= F(r )
2
2
Where:
2µE
k = 2
 2
and
 2µ 

F(r ) = 2 V (r ) ψ (r ) 
The
incident
wave
describes
a
free
particle
traveling
in
the
z
direction.
It
is
given
as

ikz
ψ inc (r ) = e €
We
also
know
that
[∇ + k ] e
ikz
[∇ + k ] e
ikz
2
2
= 0
Suggested
homework:
Demonstrate
that
2
2
= 0
Therefore,
the
equations
to
be
solved
is


[ ∇ + k ] ψsc (r )= F(r ) 2
with
2
 2µ 

F(r ) = 2 V (r ) ψ (r )



Recall
that
ψ (r ) is
included
in
F(r ) €
€

Green’s
Function
ψ 0 (r ) Solution
at
the
detector
due
to
a
point
source
 
at
r − r ' €

 
[ ∇ + k ] ψ0 (r )= δ(r −r ') 2
2

ψ 0 (r ) is
given
as
 
ik|r −r '|

e
ψ 0 (r ) = −
  4 π | r − r '|

The
solution
for
ψ sc (r ) is
obtained
by
integrating
in
volume
 
ik | r −r '|
τ
€
ψ sc (r ) = − ∫
or
€
τ
e

  F( r ')dτ
4 π | r − r '|
€

µ
ψ sc ( r ) = −
2π 2
 
ik|r −r '|
∫
τ

e

  V (r ')ψ (r ') dτ
4 π | r − r '|
Nice
solution
 of
little
value
since
neither

V ( r ) nor
ψ ( r ) are
known
quantities.
Before
proceeding
further
note
that
for


| r |>>| r '|
€
 1/ 2

 
r '.uˆ 
| r − r '| ≈ r 1− 2

r  or
 

| r − r '| ≈ r − r '.uˆ where

r
uˆ =
r
Suggested
homework:
Show
that

 
| r − r '| ≈ r − r '.uˆ 
The
function
ψ sc (r ) becomes,
  µ
ψ sc ( r ) = −
2
€
π

2


−ikr '.uˆ
 e ikr


V (r ')ψ (r ') dτ 
 r

−ikr '.uˆ


V (r ')ψ (r ') dτ
∫e
τ
Defining
µ
f (θ ) = −
2π 2
∫e
τ
Hence

e ikr
ψ sc (r ) = f (θ )
r
We
know
that
the
differential
scattering
cross
section
is
given
as
d 2σ
2
σ (θ ) =
= f (θ )
dEdΩ
µ
σ (θ ) = −
2π 2
∫e

−ikr '.uˆ


V ( r ')ψ (r ') dτ
τ


ψ
(r
)
The
functions
and
V (r ) are
still
not
Known!!!
€
€
2
Born
Approximation
Assumption:

The
complete
wave
function
ψ (r ) in
the
potential
region
is
replaced
by
the
incident
wave
function,
that
is,


ikz € ikr .uˆz
ψ (r ) = e = e
Rationale:
Incident
particle
is
weakly
scattered
in
the
potential
region,
µ
σ (θ ) = −
2π 2
∫e
τ

−ikr '.(uˆz −uˆ )

V (r ') dτ
2
€
€

V (r ) is
still
not
known!!
Fermi
Pseudopotencial
Fermi
came
up
with
a
clever
idea
for
an
expression
for V ( r ) when
the
neutron
energy
is
very
low.
Assumptions:
€
(a)
The
potential
is
applied
to
s­wave,
angular
momentum
l=0,
that
is,
the
potential
is
spherically
symmetric;
(b)
Short
range
potential
The
potential
suggested
by
Fermi
is

 2π 2
V (r ) =
a δ (r )
µ
Where
a
is
the
scattering
length
To
understand
the
concept
introduced
by
Fermi
let’s
continue
examining
the
scattering
of
neutrons
by
a
single
nucleus
with
no
spin
effect
considered.
Recall
that:

e
ikz
ψ ( r ) = e + f (θ )
r
for
r > R ikr
2µE
2
k
=
Low
energy
neutron,
since
 2 ,
implies
€
that
k → 0 Hence
€
lim r ψ (r) = r + lim f (θ )
k →0
r> R
k →0
r> R
The
scattering
length
a
is
defined
as
a = − lim f (θ )
k →0
r> R
Interpretation
of
the
scattering
length
a
For
r=R
at
the
nuclear
surface
we
have
€
R ψ (R) = R − a or
a = R[1− ψ ( R)] (a)
ψ (R) = 0 impenetrable
sphere
and
a=R
€
(b)
ψ (R) < 1 a
>
0
and
a
<
R
€
Realistic
potential
indicating
bound
states
(c)
ψ ( R) > 1
a
<
0
€
No
bound
states!!
The
differential
scattering
cross
section
with
the
Fermi
pseudopotential
is
then
µ
σ (θ ) = −
2π 2
∫e

−ikr .(uˆz −uˆ )
τ
2π

a δ (r ) dτ
µ
for
k → 0 (low
energy
neutrons)
€
σ (θ ) = a
2
The
total
scattering
cross
section
is
σs =
or
∫ σ (θ ) dΩ s
4π
σ s = 4π a
2
2
2
The
derivation
so
far
assumes
that
the
incident
neutrons
have
very
low
energy.
It
is
also
assumed
that
the
nucleus
of
mass
M
is
not
bound;
therefore,
the
scattering
length
is
called
afree.
If
the
neutron
interacts
with
a
molecule
of
mass
M’
in
which
the
nucleus
of
mass
M
is
included,
i.e.,
nucleus
bound
to
the
molecule,
the
Schrödinger
equation
for
the
system
neutron‐molecule
will
include
the
mass
between
the
neutron
(m)
and
the
molecule
(M’).
The
scenarios
are:
(a) System
m
and
M
 2 2

 
− ∇ + V ( r ) ψ ( r )=Eψ (r )
 2µ

where
µ=
mM
m+ M
€
€
(b) System
m
and
M’
 2 2

 
∇ + V ( r ) ψ ( r )=Eψ (r )
−
 2µ'

where
µ' =
mM '
m + M' The
Schrödinger
equation
for
the
system
m
and
M’
can
be
written
as
 2 2  µ'   µ'

− ∇ ψ (r )+ V (r )ψ (r )= Eψ (r )
µ
µ
2µ
We
see
an
effective
potential
for
the
system
m ‐ M ’
a s
 µ' 
Veff (r ) = V (r )
µ
As
we
have
seen,
in
the
Born
approximation
together
with
the
Fermi
pseudopotential
the
scattering
length
with
the
effective
potential
is
abound
or
abound
µ
=
2
2π

∫ Veff (r )dτ
τ
m+ M µ
=
2
M 2π

∫ V (r )dτ
τ
The
relation
between
abound
and
afree
is
abound
m+ M
=
a free
M
Defining
A
as
the
ratio
between
the
nucleus
mass
M
and
the
neutron
mass
m,
A=M/m
we
have
abound
A +1
a free
=
A
Since
the
low‐energy
bound‐scattering
cross‐
section
is
bound
2
bound s
σ
= 4π a
The
relation
between
the
bound
and
unbound
scattering
cross
section
is
2
σ
bound
s
 A + 1
=
 σ free
 A 
As
an
example,
for
A=1
(hydrogen)
σ sbound = 4 σ free Suggested
homework:
(a) Which
one
is
measured
σ sfree ?
(b) If
both,
how?
€
€
σ sbound or
MIT OpenCourseWare
http://ocw.mit.edu
22.106 Neutron Interactions and Applications
Spring 2010
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