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.
€
Scattering:
Coherent
and
Incoherent
(Interference
effects)
(A) Coherent
Elastic
Scattering
1. Occurs
for
waves
describing
incident
neutrons
having
the
same
energies
and
the
same
spin
state
2.
For
a
incident
neutron
characterized
by
e ikz each
nucleus
in
the
target
emits
a
scattered
wave
(s­wave,
low
energy)
€
ψ si = −ai
e
iki ri
ri
(Relative
coordinate
systems
attached
to
the
ith
nucleus)
3. For
N
nuclei
in
the
target,
with
same
energy
and
spin
in
the
laboratory
system,
the
total
wave
(added
coherently)
is
N
! !
iki r #R i
e
" s = $ #ai ! !
r # Ri
i=1
!
!
Ri Laboratory
position
of
the
ith
nucleus
4. Coherent
scattering
affects
only
the
!
angular
distribution
of
scattered
neutrons.
It
does
not
affect
the
total
scattering
cross
section
(neutrons
are
conserved
in
the
scattering)
€
Simple
System
The
scattered
waves
arriving
at
the
detector
are
combinations
of
the
waves
scattered
at
A
and
B
and
are
give
by
e ikr
e ik(r +BC )
ψ sc (r,θ,ϕ ) → −a
−a
r + BC r
The
scattering
lengths
are
the
same
for
both
nuclei.
€
Since
r >> BC ,
r + BC in
the
denominator
can
be
replaced
by
r
BC is
give
as
€ The
segment
€
€ θ cos ϕ
BC = d sin
Suggested
Homework:
Show
the
above
expression
Hint:
Calculate
the
angle
between
two
vectors
in
spherical
coordinates.
The
total
scattered‐wave
function
is
e ikr
ikd sin θ cosϕ
ψ sc (r,θ,ϕ ) = −
a
+
ae
[
]
r
Defining
the
scattering
function
as
f (θ,ϕ ) = [ a + a e ikd sin θ cosϕ ] The
differential
scattering
cross
section
is
σ s (θ,ϕ ) = f (θ,ϕ )
2
The
differential
scattering
cross
section
per
nucleus
for
our
simple
system
is
obtained
by
using
the
expression
for
f (θ,ϕ ) with
a
factor
½
(two
nuclei)
1 2
σ s (θ,ϕ ) = €a [1+ cos(kd sin θ cosϕ )]
2
Suggested
homework:
Behavior
of
σ s (θ,ϕ ) for
kd
<<
1
(low
energy)
and
kd
>>
1,
(high
energy)
€
Consider
the
special
case
for
ϕ = 0 ,
i.
e.,
cosϕ = 1
1 2
σ s (θ,0) = a [1+ cos(kd
sin θ )]
€
2
€
Since
k =
€
2π
λ 
1 2
d
σ s (θ,0) = a 1+ cos(2π sin θ )

2 
λ
We
see
that
the
maximum
of
σ s (θ,ϕ ) occurs
when
nλ = d sin θ
n = 1,2,3... €
This
is
the
Bragg’s
law
for
our
simple
system.
In
general
the
situation
is:
Neutrons
incident
are
scattered
by
an
angle
θ Bragg’s
Law
is
nλ = 2d sin θ
n = 1,2,3... €
Note
that
for:
(a) λ > 2d no
Bragg
scattering
is
possible
(b) Maximum
occur
for
λ = 2dmax ;
this
is
the
Bragg
cutoff
€
We
know
that
€
2.86 ×10−9
λ=
E(eV )
(cm)
Suggested
homework:
Show
that
λ = f ( E ) as
above
The
expression
for
the
Bragg
cutoff
is
€
€
dmax
0.143
=
E(eV ) (Å)
1Å
= 10−8 cm €
€
Calculate
dmax = ? €
(B) Incoherent
Elastic
Scattering:
1. Scattering
by
non
identical
nuclei
2. Nuclear
spin
3. Non‐constant
nuclear
spacing
(Crystal)
(a)
Non­identical
nuclei
(continuing
with
the
simple
system)
In
the
simple
system
we
have
been
studying
for
the
scattering
of
two
non‐identical
nuclei
there
will
two
distinct
scattering
lengths a1 and
a2 The
differential
scattering
cross
section
is
€
€
1
ikd sin θ cosϕ
σ s (θ,ϕ ) = a1 + a2e
2
2
or
1 2
σ s (θ,ϕ ) = [ a1 + a22 + 2a1a2 cos(kd sin θ cos ϕ )]
2
Maximum:
σ
max
s
1
(θ,ϕ ) = (a1 + a2 ) 2 2
Minimum:
€
σ
€
€
€
min
s
1
(θ,ϕ ) = (a1 − a2 ) 2 2
If
only
coherent
scattering
is
present
a1 = a2 min
and
σ s (θ,ϕ ) = 0 .
min
σ
The
nonzero
s (θ,ϕ ) = 0 indicates
the
presence
of
incoherent
sca€
ttering.
€
In
general
for
a
number
N
of
nuclei
with
scattering
lengths
an ( n = 1, 2, 3,...,N ) the
differential
scattering
cross
section
per
nuclei
is:
€
2
1
σ s (θ,ϕ ) =
N
N
∑ a G (k,R )
n
n
n
n=1
Rn Position
of
the
nth
nucleus
€
€
€
€
€
*
If
Gn is
the
complex
conjugate
of
Gn then
Gn Gn* = Gn*Gn = 1
Some
definitions:
€
1
< a >=
N
N
∑a
n=1
n
€
€
The
expression
for
σ s (θ,ϕ ) can
be
written
as
1
σ s (θ,ϕ ) =
€N
N
N
∑ ∑a a
n
*
G
G
m n m
n=1 m=1
By
using
an = ( an − < a >)+ < a > we
have
2
2
2 1
σ s (θ,ϕ ) =< a > − < a > + < a >
N
Suggested
homework:
Show
the
expression
above
N
∑a G
n
n=1
2
n
The
scattering
cross
section
is
σs =
Then
∫ σ (θ,ϕ )dΩ s
4π
σ s = 4 π (< a 2 > − < a > 2 ) + 4 π < a > 2
€
The
coherent
and
incoherent
cross
sections
are:
i. Coherent
ii. Incoherent
incoh
s
σ
€
€
σ scoh = 4 π < a > 2 2
2
= 4 π (< a > − < a > )
(b) Nuclear
spin
Why
is
this
a
cause
for
interference??
Potential
neutron‐nucleus
depends
on
the
orientation
of
the
spin
vector.
 
V (r) ∝ I . i 
I → nuclear spin

i → neutron spin (1/2) €
€
€



Channel
spin:
J is
a
sum
of
I and
i that
is
  
J=I+i
Therefo
€re:
€
€
J + = I + 1/ 2
and
J − = I −1/2 €
The
statistical
spin
factor
is
given
as
2J + 1
g(J) =
(2i + 1)(2I + 1) For
each
J + and
J − we
have
I
g = €
€
2I + 1 −
and
I +1
g =
2I + 1 +
The
spin
coherent
and
incoherent
cross
sections
are
i. Spin
coherent
σ
coh
s
+
2
+
−
2
− = 4π (g a + g a )
ii. Spin
incoherent
σ
incoh
s
−
+
= 4 πg g (a+ − a− )
2
Example
for
1H
€
a+ = −24.0 fm
a+ = 5.3 fm
€
€
−13
1
fm
=
10
cm and
I
where
Calculate:
€
coh
σs =?
= 1/2 €σ incoh = ?
s
σs = ?
Inelastic
Effects
In
the
“Chemical
energy
regions”
the
notion
of
elastic
and
inelastic
effects
are
related
to
the
energy
levels
of
the
target.
No
change
in
the
energy
levels
leads
to
an
elastic
scattering,
otherwise,
it
is
inelastic.
Let’s
examine
the
simple
case
of
a
diatomic
molecular
(Energy
Levels).
The
possibilities
are:
Translation,
Vibration
and
Rotation.
€
  
r = r1 − r2 and
 ' '
r = r1 − r2 The
reduce
mass
is:
m1m2
µ=
m1 + m2 (a) Vibrational
mode:
Assume
the
mass
m1 and
m2 are
bound
together
by
an
elastic
spring
such
as
F = −k(x
€ − x 0€) Classical
Quantum − Mechanics
d2x
= −k(x − x 0 )
2
dt
x = x 0 sin(wt + ϕ )
d 2ψ 2m
1 2
kx )ψ = 0
(E
+
−
2
2
dx

2
w=
k
;
m
w = 2πν
€
The
energy
levels
will
be
E 0 = 1 w
2
3
E1 = w
2
5
E
2 = w
2
etc
€
€
1
E n = w(n + ); n = 0,1,2,...
2
€
€
(b)
Rotational
Mode:
(i)
Classical
energy
of
the
system
1
1
2
E = m1υ1 + m2υ 22
2
2
for
υ1 = wr1
then
and υ 2 = wr2 1
E = I w2
2
where I = m1r12 + m2 r22
Using
€
  
r = r1 + r2
and
m1r1 + m2 r2 = 0
The
problem
is
reduced
to
a
one‐body
problem
€
The
angular
momentum
L = mrυ
or L = mr 2 w or L = Iw
The rotation energy becomes
L2
E=
2I
(ii)
Quantum‐mechanics
viewpoint
L =  l( l + 1) the energy will be :
2
€
2

E = l(l + 1) where l = 0,1, 2,...
2I
2
Vibrational­Rotational
Energy
Levels
€
In
the
inelastic
scattering
process
the
vibration
mode
of
the
system
(a
crystal
for
instance)
may
change
due
to
a
collision
with
neutrons.
The
vibrational
energy
quantized
is
called
a
phonon.
In
a
more
realistic
system
where
several
nuclei
are
present
it
is
not
possible
to
identify
individual
phonon.
In
that
case
one
uses
the
concept
of
phonon
spectrum
(phonon
distribution)
usually
a
function
of
the
frequency
w (2πν ). The
energy
of
a
phonon
of
frequency
w is
w and
the
energy
spectrum
is
f ( w ) €
€
€
MIT OpenCourseWare
http://ocw.mit.edu
22.106 Neutron Interactions and Applications
Spring 2010
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