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##

fusion

Protons from
fusion in the deuterium layer can cause a background
in the
measurements depending on the energy cut (recall that the
proton energy is 3 MeV). They can come from two different sources: (a)
direct stopping of muons in the deuterium layer, or (b) recycled muons,
*i.e.*, the muons released after the fusion reaction (with the
probability
,
where
is the sticking
probability). For both cases, because of solid state effects and finite
thickness, estimating the proton yield is difficult, and there is yet no
satisfactory theoretical model. The use of the Monte Carlo is untested in
these conditions, and its reliability is questionable without solid state
effects included in the input.

Nonetheless, we first make an analytical estimate using the two node
kinetics model which successfully, if accidentally, described the *time*
evolution of
fusion in a
bulk solid deuterium [199]. According to this kinetics
approximation, with an assumption of an infinite target and ignoring
cycling,
fusion yield per muon
can be obtained as:

where
is the target density in units of liquid
hydrogen atomic density,
are the initial hyperfine populations, and
s^{-1} is the free muon decay rate. Using the
*effective* formation rates from our measurements in thick solid
deuterium targets [199,166], *i.e.*, molecular formation
rates from each hyperfine state,
s^{-1},
s^{-1}, and
the spin flip rate,
s, we
obtain
.
The fusion branching ratio into
protons
is a somewhat complicated function of atomic hyperfine
states and molecular angular momentum states, but for our analysis here, it
suffices to let
.
Hence we have the proton yield per
muon in the kinetics model,

of which 30% comes from the fast (34 s^{-1}) component, and the
rest with the slower time slope.
This level of proton yield would give a non-negligible contribution in the
yield measurements,
when the direct stops and the recycling components are combined. However, a
proton yield at the 10% level relative to
can be excluded by
comparing the expected peak shape, simulated with an energy loss Monte Carlo
program [212], with experimental energy spectra. Recall that
the *dd* proton, with an energy of 3 MeV, has energy loss much smaller than
a 3.5 MeV ,
by a factor of roughly
,
hence a 10%
proton yield would appear as a sharp peak in the Si energy spectra, which
is absent from the data.
Given the inconsistency with this simple analytical approach, we pay closer
attention to our data set to extract the proton contribution in the
sections which follow.

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