The density of CR-39 is 1.31 g/cm^3.
Mean free path length L, given rho the cross section, and n the
number of particles per unit volume, is:
L = 1/(rho*n)
CR-39 is C12H18O7, molecular weight 274 a.m.u. It is therefore
(12*12)/274 = 52.5% carbon by weight. The carbon density is 0.525*
(1.31 g/cm^3) = 0.688 g/cm^3. At 12 g/mol, the carbon density is
(0.688 g/cm^3)/(12 g/mol) = 0.05733 mol/cm^3, or (Navrogadro atoms/
mol)*(0.05733 mol/cm^3) = (6.022x10^23 atoms/mol)*(0.05733 mol/cm^3)
= 3.452x10^22 atoms/cm^3.
A barn is 10^-28 m^2. Assuming the neutron came from:
D + T --> 4He (3.5 MeV) + n (14.1 MeV)
it is 14.1 MeV, so we can assume the cross section of the 12C(n,n’)
3alpha is 0.3 barns = 3x10^-29 m^2.
The mean free path of neutrons in carbon of that density is:
L = 1/((3x10^-29 m^2)*(3.452x10^22 atoms/cm^3) = 0.965 m
Using the Beer-Lambert law, we have the beam intensity I, giving
initial beam intensity I0:
I = I0 * EXP(-x/L)
and thus the fraction R of neutrons stopped in a slab of thickness x
of 0.1 cm:
R = 1 - I/I0 = 1 - EXP(-x/L) = 1 - EXP(-0.001 m)/(0.965 m) = 1 - .
9989642 = 1.036x10^-3
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. 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
6000 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 12000 neutrons. Given a rate of 10 per 2
weeks or 1 per 1.2096x10^6 seconds detected. The actual high energy
neutron rate is thus roughly one every 100 seconds. Neutron
production rate is probably way less than that because the estimate
of 100 times attenuation for O and H is probably way too high.
One neutron in 100 seconds or more is not a very big number. It is a
credible number, assuming my calculations are correct. It would take
a very good neutron spectrometer and laboratory conditions to
distinguish that from background. OTOH, if trace tritium doping were
used that number could easily be bumped up by a factor of 10,000,
assuming the triple tracks are indeed from 12C(n,n’)3alpha reactions.
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
