Forest He-3 Tubes

Thermal neutron capture of He-3 may be represented by the reaction below

$n + He^3 \rightarrow p + H^3$

About 764 keV of energy is liberated in this nuclear reaction and distributed between the final products according to their masses.

Cons Momentum $\Rightarrow$

$m_p v_p = m_{H^3} v_{H^3} \Rightarrow v_p = 3v_{H^3}$

Because the proton is about a factor of 3 lighter than Tritium (H^3) , it will have more kinetic energy by about a factor of 3 (about 573 keV). This liberated proton can ionize other He-3 atoms via the reaction

$p+He^3 \rightarrow p + He^3(+) + e^-$

The same proton will ionize several He-3 atoms when dissipating the 573 keV kinetic energy. Once you have the creation of ions, you can construct detectors to collect and measure the electrons.

The Tritium (H^3) can also ionize the gas but due to its higher mass it does not travel as far (shorter range) as the proton and makes a smaller contribution to the ionization signal. Tritium decays to He-3 after about 12 years ( neutron is converted to proton)

$H^3 \rightarrow He^3 + e^- + \bar{\nu_e}$

reference: J. W. Leake, "Nuclear Instruments and Methods", Vol. 63, page 329, 1968).

The probability of neutron capture is measured in terms of a cross section. There is a nuclear data base for neutron capture located at LBL. The "free" neutron thermal total cross section ($\sigma$) is $3.10 \pm 0.13$ barns ( 1 barn = $10^{-24} cm^2$).

Total cross section is defined as

$\sigma \equiv \frac{\# particles\; scattered} {\frac{ \# incident \; particles}{Area}}$

$\# protons = \sigma \frac{\# incident particles}{Area}$

$= \sigma \rho \times L$

$= ( 3 \times 10^{-24} cm^2) (10 Atm) \left ( 2.7 \times 10^{19} \frac{atoms}{cm^3} \right ) (76 cm) = 0.06 partices \Rightarrow \epsilon = 6$ %

where

$\rho$ = density of the He-3 target

L = length of the target.

10-atm He-3, 2.54 cm diameter tube, 76 cm long, poly moderator, cadmium metal, Boron loaded shielding.

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