22.54 Neutron Interactions and Applications (Spring 2004)
Chapter 6 (2/24/04)
Energy Transfer Kernel F(E → E')
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References -J. R. Lamarsh, Introduction to Nuclear Reactor Theory (Addison-Wesley, Reading,
1966), chap 2.
S. Yip, 22.111 Lecture Notes (1975), chap 7.
Allan F. Henry, Nuclear-Reactor Analysis (MIT Press, 1975), chap 2.
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Having discussed the calculation of the angular differential scattering cross
section σ (θ ) = dσ / d Ω , and its integral, the scattering cross section σ ( E ) , we can now ask
about the energy differential scattering cross section dσ / dE ' . When properly normalized,
this cross section becomes the probability that given the neutron is scattered at E, it will
have post-collision energy in the interval dE' about the energy E'. A similar
normalization will give a corresponding distribution for dσ / d Ω . Since the two
differential cross sections are related in that their integrals are just the elastic scattering
cross section
σ ( E ) = ∫ d Ω ( dσ / d Ω )
= ∫ dE '( dσ / dE ')
(6.1)
we can define two probability distributions, P (Ω) and F ( E → E ') , where
P (Ω) d Ω ≡
=
probability of scattering (at energy E) into solid angle element dΩ about Ω
1
( )
dσ
σ (E) dΩ
(6.2)
dΩ
and
F ( E → E ') dE ' ≡ probability of scattering (at
=
 dσ  dE '


σ ( E )  dE ' 
1
E) into dE' about E'
(6.3)
In view of their indicated connections to σ ( E ) , it is not surprising that the two
distributions are directly related to each other such that if one is known the other is
readily obtained by a transformation. Indeed this is a general property of the
transformation of distributions. Suppose g(y) and f(x) are both distributions and y=y(x),
then g(y) can be obtained from f(x) by the relation,
1
g ( y ) dy = f ( x ) dx
(6.4)
g ( y ) = f ( x ) dx / dy
(6.5)
or
To apply this argument to the angular and energy probability distributions, (6.2)
and (6.3), we first need to reduce the former quantity which is a function of two
variables, the angles θ and ϕ , since the latter is a function of one variable. The
reduction is possible because we are dealing with central force scattering, in which case
the probability distribution P (Ω) does not depend on the azimuthal angle ϕ . To make
this explicit we write henceforth P (Ω) = P (θ ) , and integrate (6.2) over ϕ ,
∞
∫ P(θ ) sin θ dθ dϕ ≡ G (θ )dθ
(6.6)
ϕ =0
The reduced angular distribution is G (θ ) , it is only a function of the polar angle θ , or the
angle of scattering. We have previously derived a particularly simple relation between
the energy of the scattered neutron E' and the scattering angle in CMCS, see Eq.(3.10) in
Lec 3 (2003), E ' = ( E / 2){(1 + α ) + (1 − α ) cos θ c ] . This result shows that there is a one-to-one
correspondence between E' and θ c . Notice that this correspondence also holds between
E' and the scattering angle in LCS. On the other hand, for bona fide nuclear reactions
(elastic scattering is not considered proper nuclear reaction), recall that one can doublevalued solutions to the Q-equation. In the example considered in Lec 3 (2003) (p.7) we
had two different values of E' for the same angle in LCS.
Our interest here is to relate the two probability distributions, G (θ ) and
using (3.10) to evaluate the Jacobian of transformation dx / dy in (6.5). Thus
we write
F ( E → E ') ,
(6.7)
F ( E → E ') dE ' = G (θ c ) dθ c
The corresponding physical statement is that the probability of scattering into dE' about
E' is the same as the probability of scattering through an angle θ c . If the angular
distribution in CMCS were known, then the energy scattering kernel becomes
(6.8)
F ( E → E ') = G (θ c ) dθ c / dE '
From (3.10) we find
dθ c / dE ' = [( E / 2)(1 − α ) sin θ c ]
−1
(6.9)
Eqs. (6.8) and (6.9) are as far as we can go without specifying the angular distribution.
We now confine our discussions to low-energy, s-wave scattering in which case the
angular distribution in CMCS, P (Ω c ) , is spherically symmetric. This means
P (Ω ) = 1/ 4π
(6.10)
or
2
(6.11)
G (θ c ) = (1/ 2) sin θ c
Inserting (6.9) and (6.11) into (6.8) we obtain the energy transfer kernel
F ( E → E ') = 1/ E (1 − α ) ,
αE ≤ E ' ≤ E
(6.12)
0.
otherwise
Notice that the upper and lower bounds on E' correspond respectively to forward
scattering, θ c = 0, where the neutron loses no energy, and to backward scattering, θ c = π ,
where the neutron suffers maximum energy loss. The student is advised to make sure to
specify the kernel throughout the entire energy range by not forgetting to write out the
second line of (6.12) when asked to give the energy transfer kernel. A sketch of the
energy transfer kernel is shown in Fig. 1.
Fig.1. The energy transfer kernel F ( E → E ') derived under the assumptions of elastic
scattering, target nucleus at rest, and spherically symmetric scattering in CMCS, with
α = [( A − 1) /( A + 1)] and A = M/m.
2
The very simple form of F ( E → E ') makes it easy to understand all its features.
The probability distribution
is uniform in the range between (α E , E ) , where we recall
2
α = [ ( A − 1) /( A + 1) ] , with A = M/m, and zero outside this range. The existence of a cutoff
in the range of energy that can be transferred from the neutron to the target nucleus
corresponds to the range of scattering angle that one can have, minimum value of θ c is
zero which is forward scattering (no collision) and maximum is π which is backward
scattering. In the case of scattering by hydrogen, α = 0, so the energy range extends
down to zero. With neutron and hydrogen having the same mass (true only for the
purpose of our calculation) it is not surprising that in a collision the neutron can transfer
all its energy to the hydrogen. Lastly we can ask why is the probability distribution
uniform. The answer is again quite straightforward, namely the uniform distribution is a
direct consequence of the spherically symmetric form of the angular probability
distribution, which in turn arises because of s-wave scattering.
Another way to discuss F ( E → E ') is to examine the assumptions that have been
made in deriving (6.12). There are three such assumptions, (i) elastic scattering, (ii)
target nucleus at rest, and (iii) spherically symmetric scattering in CMCS. Recalling our
discussion of kinematics of nuclear reactions (Lec 3 (2003)) elastic scattering is the case
of Q = 0, and the assumption at the outset of target nucleus being at rest allows us to
simplify the algebra considerably in the subsequent analysis. Relative to our discussion
of cross section calculation (Chap 4), the conditions of elastic scattering and stationary
target are equivalent to our solution of the Schrodinger equation for an effective particle
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in CMCS. While the third assumption, spherically symmetric scattering in CMCS, has
no counterpart in any of the discussions in Lec 3 (2003), we know from Chap 4 that swave scattering is spherically symmetric. Thus, the question is when can we ignore all
the other partial-wave contributions to the scattering. In Lec 4 (2003) we have
emphasized that this would be a good approximation under the condition of low-energy
scattering, that is, kro < 1. Thus, the energy transfer kernel F ( E → E ') given in (6.12) is
valid for neutrons at sufficiently low energy. We can estimate an upper limit on the
energy by taking kro ~ 0.1, with ro ~ 2 x 10-12 cm. This gives k ~ 5 x 1011 cm-1, or E ~ 47
keV. Certainly for neutrons at thermal energies or 100 eV, or even 1 keV, (6.12) should
be applicable.
Suppose we wish to relax the assumption of spherically symmetric scattering in
CMCS. What would F ( E → E ') look like if the scattering were biased either in the
forward or in the backward direction? One can postulate simple forms of bias instead of
(6.10) and carry through the transformation as before, along with using (3.10). One
should then obtain non-uniform distributions in the final energy E'. The correspondence
between angular distribution and energy distribution is relatively simple to figure out. If
the angular distribution favors the forward scattering direction, then the energy
distribution should show a bias toward smaller energy transfer; similarly a backward
scattering bias should translate into an energy distribution that favors larger energy
transfer. Fig. 2 shows the characteristic behavior that is expected.
Fig.2. Schematic behavior of energy transfer kernel F ( E → E ') when the angular
distribution in CMCS P (Ω c ) is peaked in the forward (backward) direction. The dashed
line is the result shown in Fig. 1 for spherically symmetric scattering (isotropic in
CMCS).
What if we wish to relax the other two assumptions? Let us consider for a moment what
one can say about other neutron scattering processes besides elastic potential scattering.
There are two such processes, one is elastic resonance scattering which we have touched
on in Lec 2 (2003) (cf. (2.12)) with regard to the dependence of the scattering cross
section on the incoming neutron energy. In addition to elastic potential and resonance
scattering, neutrons with sufficient energy can induce inelastic scattering, a reaction
involving the formation of a compound nucleus which decays to an excited state of the
target nucleus with the emission of a neutron at considerably lower energy. If the need
arises, our theoretical understanding of nuclear reactions is probably good enough to
allow us to construct an energy transfer kernel for these processes. On the other hand,
such analysis, to our knowledge, would be well beyond the scope of any nuclear
engineering course that is being taught.
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In contrast to elastic scattering, relaxing the assumption of target nucleus at rest is
a very relevant issue, scientifically and technologically. When neutrons get down to
thermal energies, it is no longer justified to assume that they move much faster than the
target nucleus. In that case, target nucleus motion becomes a significant variable and one
must take this effect into account. It turns out that this not such a simple issue, we will
return to discuss it in some detail in the next lecture (Chap 7).
Before closing this lecture we take up one final topic, that of the behavior of the
angular differential cross section dσ / d Ωc = σ (θ c ) , which has played a central role in the
development in this lecture. Given its importance, it would be instructive to see an
example or two of this distribution. In Fig. 3 we show the differential cross section for a
10000
8000
1000
800
6000
600
4000
400
2000
0.5 MeV
σ (θc ), millibarns/steradian
σ (θc ), millibarns/steradian
200
100
80
60
40
14 MeV
20
10
8
1000
800
600
0.5 MeV
400
200
100
80
60
6
14 MeV
40
4
20
2
1.0
0.8
0.6
0.4
0.2
0 - 0.2 - 0.4 - 0.6 - 0.8 - 1.0
COS θ
10
1.0
0.8
0.6
0.4
0.2
0 - 0.2 - 0.4 - 0.6 - 0.8 - 1.0
COS θ
Fig. 3. Angular differential scattering cross section of C12 (a, left panel) and U238 (b,
right panel) at two incident neutron energies, 0.5 MeV and 14 MeV.
light element, C12, and a heavy element, U238, each at two incoming neutron energies,
0.5 MeV and 14 MeV. The distributions are plotted as functions of µc = cosθ c . Based on
what we have discussed above, one expects that at low energy the distribution should be
independent of the scattering angle. Indeed this is clearly seen in the result for C at 0.5
MeV. We can estimate what is kro in this case. Let ro ~1.2 x A1/3 F = 2.75 x 10-13 cm. At
0.5 MeV,
k 2 = 2mE / = 2 = 2 x1.67 x10−24 x0.5 x106 x1.6 x10−12 /10−54 = 2.6 x1024
cm-2
so kro ~ 0.44, small enough to satisfy the s-wave scattering approximation. On the other
hand, at 14 MeV, k = 8.5 x 1012 cm-1, and kro ~ 2.35. In this case one would expect that
the p-wave and possibly the other higher-order partial wave contributions to become
important. The breakdown of the s-wave approximation means that the angular cross
section should be peaked in the forward scattering direction. This is seen in Fig. 3(a).
Besides the forward bias, one should notice that the angular distribution is oscillatory.
This can be understood as evidence of neutron diffraction by the nucleons, in the same
manner as thermal neutron diffraction in a sample of atoms and molecules. When the
wavelength of the particle undergoing scattering becomes comparable to the spacing
between a collection of scattering centers, one can expect interference effects. With
thermal neutrons the wavelength is of the order of angstroms (10-8 cm) which are
comparable to intermolecular distances in condensed matter - this leads to interference
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among the scattered waves and the appearance of oscillations (the diffraction pattern). In
the present case, the neutron wavelength is λ = 2π / k = 7.39 F , apparently short enough to
begin to be sensitive to interference effects among different nucleons.
In view of the results for C12, we can readily predict what should happen in the
case of U238. Since ro is now ~ 7.44 F, we see that at 0.5 MeV, kro is now 1.66x0.744 =
1.19. It is then not surprising that the characteristic forward peaking, signaling the
breakdown of the s-wave scattering approximation, is seen in Fig. 3(b). That said, it is
also to be expected that at 14 MeV, with kro ~ 6.3, quite pronounced diffraction behavior
should be observed.
In the next lecture we will continue to discuss the energy dependence of the
scattering cross section, focusing on the 'total' cross section σ ( E ) as well as F ( E → E ')
when thermal motion and chemical binding effects come into play. Together Lectures 6
and 7 will provide the understanding of cross section behavior that will form the basis of
neutron interaction in preparation for our discussion of neutron transport.
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