Microdosimetric Basis for Exposure Limits

Abstract

Consideration of the energy deposited by ionizing radiation in microscopic volumes has led to new insights into dosimetric concepts at the levels of interest in radiation protection. Large amounts of energy are deposited by the passage of low linear-energy-transfer (LET)-charged particles through small volumes. If a typical cell nucleus is considered to be about 7 micron, at an exposure rate of 2.5 X 10(-1) C kg-1 h-1 (1 mR hr-1) from a 60Co irradiation, the average cell nucleus receives one energy deposition event every 12.5 d. Biological processes, which modify radiation damage, typically occur in a few minutes to a few hours. Thus, at occupational exposure levels it is probably the irreparable or misrepaired effects of irradiation that determine the biological consequences. One goal of dosimetry is to measure the incident radiation, making it possible to predict biological risk and set meaningful exposure limits. These measurements must relate to the energy depositions that are responsible for radiation effects at low dose rates, yet the dosimetry system must not be excessively complex to allow use by operational health physicists. Furthermore, our description of the irradiation should be directly measurable. The use of quality factors based upon the energy deposition in a 1-micron-diameter volume of tissue as prescribed in International Commission on Radiation Units Report No. 40 (Joint Task Group 1986) satisfies these requirements. Instrumentation based upon measurement of lineal energy has already been successfully used in health-physics applications. Future changes in the quality factor can be accommodated by changing the algorithm in these microprocessor-based instruments.

Evidence for dose and dose rate effects in human and animal radiation studies

“…alpha particles) because, in general, the dose response for tumour induction following exposure to these types of radiation is linear, with no variation in effect with dose rate (ICRP, 1991; UNSCEAR, 1993). The reason for this may be connected to the fact that, at the level of the cell nucleus, there is no such thing as a low dose rate; although only approximately 1.5% of cells have a neutron traversal of the cell nucleus following a tissue dose of 3 mGy (30 mSv), every cell nucleus that is traversed by a neutron receives a large dose of approximately 0.2 Gy (2 Sv) (Brackenbush and Braby, 1988). Therefore, the instantaneous dose rate at the level of the cell nucleus is unavoidably high (>5 mGy h –1 (Wakeford and Tawn, 2010).…”

Biological Effects of Stellar Collapse Neutrinos

“…The study of how different patterns of spatial and temporal distributions of energy at the subcellular level affect the biological response of cells to very low doses of ionizing radiation has become a very active area of research. This is due to the increasing awareness that high linear-energy-transfer (LET) radiations (alpha particles, fission neutrons, ions) may be responsible for unique biological effects [13][14][15]. Plainly stated, the (unrestricted) LET is the amount of energy dissipated by a radiation per unit path length.…”

“…The greater effectiveness of the high-LET tracks more than compensates for their smaller number per unit dose. This is illustrated by the microdosimetric concept of Specific Energy, z = ε / m, where ε is the energy imparted by radiation to matter of mass m [13]. It is a stochastic quantity and depends on the number of energy deposition events in the mass of interest, and therefore on the ionization density of the tracks.…”

“…It is a stochastic quantity and depends on the number of energy deposition events in the mass of interest, and therefore on the ionization density of the tracks. Considering a 7 µm diameter sphere as representative of an eukaryotic cell nucleus, single 60 Co events have a mean z of 1.25 mGy, while for fission neutrons z ~ 200 mGy [13]. For an 80 keV oxygen ion, representative of stellar collapse damage (see…”

“…nm length cylinder), z = 5.5 ⋅ 10 5 Gy, 22.1 ⋅ 10 5 Gy and 56.0 ⋅ 10 5 Gy, respectively. The utility of z is illustrated as follows: for a neutron exposure (dose) of D= 3 mGy, only D / z = (3 mGy / 200 mGy) = 1.5 % of cell nuclei in the irradiated tissue will receive any damage, while each of them gets the same amount as cells exposed to 200 mGy in a dose-response study[13]. In our case, the exposure can be approximated (using the prevalent Oxygen recoils and Eq.…”