Fission Product Yield Data for the Transmutation of Minor Actinide ...

FIG. 4.2.36. n T(A) for CF249T. FIG. 4.2.37. n T (A) for TH232F.
very asymmetric mass division. As well as the peak
at symmetry that results from the assumption of
four or more neutrons being emitted near
symmetry, there are small dips on either side of the
peak, as can be seen from the circles in Figs 4.2.32–
4.2.37. The dips occur near A H = 130 and in the
complementary light product range near A = 104 for
U235T, and may be associated with condensed
scission configurations, possibly related to the
proximity of Z and N of the heavy fragment to the
spherical Z = 50 and N = 82 shells. High fragment
kinetic energies should result and have been
observed [4.2.28], and therefore excitation energies
and neutron emission are small.
The dips in the n T(A) function are present for
the fission reactions of uranium, plutonium and
americium isotopes that were investigated, but are
smaller or absent for the fission reactions investigated
for the thorium and californium isotopes. This
may be due, at least in part, to the somewhat fewer
experimental mass yield values for these reactions,
since dips are absent when model mass yields
calculated from smooth functions are used for
U235T (see Fig. 4.2.38).
Other structure appears for very asymmetric
mass division (see Figs 4.2.32–4.2.37), but the effects
vary in position, magnitude and sign for the various
fission reactions investigated, and are attributed to
the method applied to low and possibly uncertain
fission yield values that form the wings of mass yield
curves.
4.2.4.4. Model for n T(A)
n T (A) values from the Y(A) summation calculations
(circles in Figs 4.2.32–4.2.37) can be approximately
represented by the solid lines through the
FIG. 4.2.38. n T (A) for U235T, Model Y(A) values.
addition of three Gaussian functions to a constant
NT (Eq. (4.2.22)). The ECOR term serves to damp
out the peak and dips at high energies (PE), where
single particle effects are less important.
n T (A) = NT + ECOR[n S (A) + n H (A) + n L (A)]
–0.05 PE
ECOR = e
For the peak at symmetry:
n ( A) = I [ e P P]
/
pC
S P
2 /
( A-S ) C
(4.2.22)
(4.2.22a)
(4.2.22b)
I P = 20.0, = 0.0 for SF (4.2.22c)
C P = 2.0(6.0 2 + 1/12) = 72.17 (4.2.22d)
S P = (A F – n s )/2.0, n s = 4.0 or
(NYT + 1.0) if >4.0 (4.2.22e)
P
141