On Dec 7, 2009, at 1:32 PM, Jones Beene 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.
I think that is the energy required to *see* the triple tracks.
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.
To do
calculation correctly, it would seem that the cross-section for all
elastic
collisions must be considered first to take this into account. 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?
What am I missing?
You are missing this:
I wrote:
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
In 1/16 inch, or 1.5875x10^-3 m we have:
R = 1 - I/I0 = 1 - EXP(-x/L) = EXP(-0.0015875 m)/(0.965 m) = 1 - .
99835627 = 1.0644x10^-3
or 1 in about 600 neutrons.
If we assume 100 times the beam attenuation for O and H, about 1 in
60,000 neutrons in a 12C(n,n’)3alpha reaction. Given the surface area
is close to the wire, we could assume almost 50% detection rate. That
means we can see about 1 in 120,00 neutrons.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
end quote
Note the "120,00" above should of course be "120,000".
The point is I provided a fudge factor of *100 times* attenuation due
to proton and oxygen collisions. It is rough, but plenty good enough
for this purpose. The actual cross section for incoherent neutron-
hydrogen (knock ons) is about 15 times that for the 12C(n,n’)3alpha
reaction, and the coherent reaction cross section for hydrogen plus
oxygen is about equal to that of the 3 alpha reaction, at least at
the 14.1 MeV neutron wavelength, which is 7.54x10^-15 m, or
7.54x10^-5 Å. That means the "straight through" neutrons to the 12C
represent about 1/(1+15) = 1/16 = 0.0625 of the neutrons, or about
one in 16, *not* 1 in a 100, which is what I used for a fudge factor.
I gave a very conservative calculation AFAIK.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
