4 ms·

My understanding is that penetration depends on the energy associated with the the β particle [0] and tritium decay releases relatively lower-energy β particles

My understanding is that penetration depends on the energy associated with the the β particle [0] and tritium decay releases relatively lower-energy β particles [1]. The nonprofit isn't wrong... they're just saying "a .22 bullet can kill you" while parent is saying "tritium is like throwing the bullet instead of firing." [1] https://sciencedemonstrations.fas.harvard.edu/presentations/%CE%B1-%CE%B2-%CE%B3-penetration-and-shielding https://sciencedemonstrations.fas.harvard.edu/presentations/... [1] https://en.wikipedia.org/wiki/Tritium https://en.wikipedia.org/wiki/Tritium

rnhmjoj 6y ago
Exactly, the electron produced by the decay of tritium have an energy that is distributed between 0 and 18.59 keV (the Q value of the reaction), with an average of 5.68 keV (this is pretty low compared to other common radiation sources). The rest of energy is shared between the neutrino (which practically never interacts with matter) and the residual nucleus (which is very massive compared to the electron and thus can't travel far). So, electrons are the only source of concern here. Electrons can deposit energy in two ways: one is by bremsstrahlung, which is basically EM radiation (X rays, usually) emitted during a fast deceleration caused by another ion; or by ionising neutral atoms and breaking bonds. Bremsstrahlung is only significant at much higher energies, so this leaves ionisation. The energy lost per unit length (-dE/dx) depends on the electron energy and can be computed using the Bethe-Bloch formula. Integrating this quantity yields the particle range R = ∫_E₀ ⁰dE (dE/dx)⁻¹. An empirical formula that can be derived from this is the following: R = 1/ρ 0.44cm E^(1.265 - 0.0954 log(E)) where ρ is the medium density (in g/cm³) and E the electron energy (in MeV). In water you get: R = 4 μm for the average energy and 50 μm for the maximum energy, which is extremely rare, by the way. Note that to damage tissues, radiation must reach the radiosensitive layer of the skin at depths greater that 40 μm.

rnhmjoj 6y ago

Exactly, the electron produced by the decay of tritium have an energy that is distributed between 0 and 18.59 keV (the Q value of the reaction), with an average of 5.68 keV (this is pretty low compared to other common radiation sources). The rest of energy is shared between the neutrino (which practically never interacts with matter) and the residual nucleus (which is very massive compared to the electron and thus can't travel far). So, electrons are the only source of concern here. Electrons can deposit energy in two ways: one is by bremsstrahlung, which is basically EM radiation (X rays, usually) emitted during a fast deceleration caused by another ion; or by ionising neutral atoms and breaking bonds. Bremsstrahlung is only significant at much higher energies, so this leaves ionisation. The energy lost per unit length (-dE/dx) depends on the electron energy and can be computed using the Bethe-Bloch formula. Integrating this quantity yields the particle range R = ∫_E₀ ⁰dE (dE/dx)⁻¹. An empirical formula that can be derived from this is the following: R = 1/ρ 0.44cm E^(1.265 - 0.0954 log(E)) where ρ is the medium density (in g/cm³) and E the electron energy (in MeV). In water you get: R = 4 μm for the average energy and 50 μm for the maximum energy, which is extremely rare, by the way. Note that to damage tissues, radiation must reach the radiosensitive layer of the skin at depths greater that 40 μm.