Display MoreMy analysis was based on the plots on page 3 of Stelson, showing the
counts per micro Coulomb of beam current. The count data for Ni isotopes
are almost four orders of magnitude lower than for Se (which is not
present in the reactor).
In the Glowstick experiments to date, we are seeing net power (if any)
of less than 100 watts. My conclusion was meant to show that in the
Glowstick experiments, this reaction signal would not have been
measurable. If we're able to generate even 1 kW in a Glowstick, such a
signal would be possible to detect, and we'll watch for it. In such an
event, there would be lots of other signals as well, and possibly a
melt-down of reactor components.
There is not a factor 104 between 58Ni and Se, more like a factor of
20. But let's forget about Se, I don't know why that got into the
discussion.
I find your comments defensive and negative. If p+7Li is important in a
LENR reactor, measuring gammas would be the ultimate proof that there is
a reaction and the gamma yield would give a handle on the excess
energy. That would convince even the hardcore nuclear physicists that
you have got something! It would definitely be a good signal. Even a
negative result would be of interest since it would give a rather
stringent upper limit of the p+7Li contribution.
We may be willing to measure a alpha+Ni gamma spectrum at our accelerator with
the 2011 Rossi fuel samples (thick target yields, as in Stelson).
Lets calculate an estimate of the count rate of the 1454.2 keV
58Ni(2+->0+) transition. (You can renormalise the final result if you
prefer other assumptions.)
Reaction: p+7Li-->2 alpha Q=17.347 MeV.
Alpha energy = 17.347/2=8.67 MeV (all excess energy goes into kinetic energy of the alphas).
Excess energy from the reaction above: 100 W.
Number of alphas per second: 100/(8.67*106*1.6*10-19)=100*1013/(8.67*1.6)=7.2*1013 /s.
Equivalent current: I=N*q*e=7.2*1013*2*1.6*10-19*106=23 microA
So we have 23 microC per second.
The yield of 58Ni was 9.38*104 (Stelson, table 1, 7% uncertainty, E=8.013 MeV, close to alpha energy above).
Correcting for natural abundance of 58Ni: (0.68/0.996)*9.38*104= 64000.
Total activity (only 58Ni) is 23*64000=15*105 /s=15*105/(3.7*104)=40 microCi,
which is much more than a standard source (1 microCi). There is
obviously reason to take safety precautions.
A conservative photo peak efficiency of 1% would yield 15000 counts/s. That is probably more than the detector can handle!
I realize that this is, if I have the sums right, as popular as the proposal to demote Pluto to a dwarf planet!