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
σ θ d σ / d Ω , and its integral, the scattering cross section about the energy differential scattering cross section d σ / dE '
σ E , we can now ask
. 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 d σ / d Ω . Since the two differential cross sections are related in that their integrals are just the elastic scattering cross section
σ ( )
∫ d Ω ( d σ / d Ω )
= ∫ dE '( d σ / dE ') (6.1) we can define two probability distributions, P ( ) and F ( E → E ') , where
P ( ) d
probability of scattering (at energy E) into solid angle element d Ω about Ω
=
1 d
(
σ
) d Ω
σ ( ) d Ω
(6.2) and
F ( E → E ') dE ' ≡ probability of scattering (at E) into dE' about E'
=
σ
1
 d σ dE '
 dE ' (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,
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= f ( ) (6.4) or g y f ( x ) dx / dy (6.5)
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 ( ) this explicit we write henceforth does not depend on the azimuthal angle ϕ . To make
P ( ) P ( θ ) , and integrate (6.2) over ϕ ,
∞
∫ ϕ = 0
P θ d
≡
G ( θ θ (6.6)
The reduced angular distribution is G ( ) , 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
= − α θ c
] Lec 3 (2003), E ' ( E / 2){(1 + α ) + (1 . This result shows that there is a one-to-one correspondence between E' and θ . Notice that this correspondence also holds between c
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 ( )
F ( E → E ') , using (3.10) to evaluate the Jacobian of transformation / we write and
in (6.5). Thus
F ( E → E ') dE ' = G ( θ c
) d θ c
(6.7)
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 . If the angular c distribution in CMCS were known, then the energy scattering kernel becomes
F ( E → E ') = G ( θ c
) d θ c
/ dE ' (6.8)
From (3.10) we find d θ 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 ( ) 1/ 4 π (6.10) or
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G θ = (1/ 2)sin (6.11)
Inserting (6.9) and (6.11) into (6.8) we obtain the energy transfer kernel
→ E ') = 1/ E (1 − α ) , α ≤ E ' ≤ E
(6.12) otherwise
Notice that the upper and lower bounds on E' correspond respectively to forward scattering, θ = 0, where the neutron loses no energy, and to backward scattering, θ = π , c c 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. energy transfer kernel is shown in Fig. 1.
A sketch of the
Fig.1
. The energy transfer kernel
α = [( A − 1) /( A + 1)] 2
( → E ') scattering, target nucleus at rest, and spherically symmetric scattering in CMCS, with
and A = M/m.
derived under the assumptions of elastic
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
α =
[
( A − 1) /( A + 1)
]
2 , 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, kr o
< 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 kr be applicable. o
~ 0.1, with r o
~ 2 x 10 -12 cm. This gives k ~ 5 x 10 11 cm -1 , or E ~ 47 keV. Certainly for neutrons at thermal energies or 100 eV, or even 1 keV, (6.12) should
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 d σ / d Ω = σ θ )
, which has played a central role in the c c 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
Fig. 3.
Angular differential scattering cross section of C 12 (a, left panel) and U 238 (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 µ c
= cos θ 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 kr
0.5 MeV, o
in this case. Let r o
~1.2 x A 1/3 F = 2.75 x 10 -13 cm. At k 2 = 2 mE / = 2 = x x − 24 x x 6 x x − 54 = x 24 cm -2 so kr o
~ 0.44, small enough to satisfy the s-wave scattering approximation. On the other hand, at 14 MeV, k = 8.5 x 10 12 cm -1 , and kr o
~ 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 = 2 π / k = 7.39 F
, apparently short enough to begin to be sensitive to interference effects among different nucleons.
In view of the results for C case of U 238 . Since r o
12 , we can readily predict what should happen in the
is now ~ 7.44 F, we see that at 0.5 MeV, kr also to be expected that at 14 MeV, with kr o should be observed. o
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
~ 6.3, quite pronounced diffraction behavior
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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