At 05:32 PM 12/7/2009, you wrote:
-----Original Message-----
From: Horace Heffner
> Mr. Heffner has kindly done the calculations, and I wouldn't
> consider myself qualified to do them right, at least not the first
> time!
I *never* get anything right the first time. 8^)
Whoa. What about elastic collision cross sections? If there is a mistake in
your underlying assumptions - then one of them could be this. Al-Najjar et
al. (1986) reported that the threshold energy of the neutron required to
fission a carbon atom (three alpha) is 9.6 MeV. The authors apparently
agree, and cite this reference several times.
Yes. So? 9.6 MeV neutrons, at least. Horace did his calculations with 14.1.
This is a cutoff, a threshold below which - nada - so if there is an elastic
collision of the high energy neutron first, with any hydrogen atom which is
half the atoms, then the neutron is lost for further consideration.
Yes. However, the cross-section for those collisions is still small.
And most of those collisions would not take the neutron below the 9.6
MeV threshhold. In any case, moot. triple tracks indicate energetic
neutrons, there isn't any other reaction known to do that in this environment.
To do
calculation correctly, it would seem that the cross-section for all elastic
collisions must be considered first to take this into account.
It's a small factor. Most of the neutrons would not have experienced
an elastic collision.
This seems to
absent since when you say "Thus 1 in 1/(1.036x10^-3), i.e. 1 in 966 neutrons
that is not stopped by O or H is involved in a 12C(n,n')3alpha reaction
within 1 mm" but that seems to assume that there is no elastic collision
possible with the carbon. Is that true?
There could be elastic collision with carbon. That would only be seen
as a single track. And the cross section that was used was for 12C
breakup. The cross section of 0.3 barns accounts for all the other
reactions and effects that might happen, already. If I understand correctly.
A more likely source of underestimation of neutrons would be that
most of the carbon fractures would not be visible, because they would
not penetrate to the back side of the CR-39, only ones fairly close
to the back would make it. I don't recall the depth of CR-39 that the
alphas can penetrate, but it is way short of the thickness of the CR-39.
I was just reading that one can clean up the background from CR-39 by
a heavy pre-etch. That wouldn't get rid of background from neutrons,
but it would get rid of the charged particle radiation, one just
takes off the top layer, maybe 20 microns or so?
What am I missing?
Is anything loose?
