Exposure from technetium-99

The approach to this calculation can be found in several textbooks, but the one I used for this question is Principles of Radiological Health and Safety by James E. Martin and Chul Lee (Wiley – Interscience, John Wiley & Sons, Inc., Hoboken, New Jersey, 2003).

The best approach is to attack this problem in two steps. First, one should calculate the beta fluence rate at the point of interest. A second calculation is necessary to determine the beta absorption coefficient for the medium in which the exposure rate or dose rate is required. Empirical equations are available for air, tissue, and any other medium. The absorption coefficients are (from the above reference):

µ_{ß, air} = 16(E_{ß max} – 0.036)^-1.4

µ_{ß, tissue} = 18.6(E_{ß max} – 0.036)^-1.37

µ_{ß, i}  = 17(E_{ß max} )^-1.14  where i represents any other medium.

In all these equations above, the units on m_ß are expressed as the “density thickness,” i.e., cm² g^-1 and E_{ß max} has units of MeV.

The beta fluence rate (f_ß) at a specified distance can be calculated for a point source using the inverse square law.  The dose rate for a beta emitter is:

D_ß (Gy h^-1) = k (f_ß ß cm^-2 s)(E_max MeV ß^-1)(µ_{ß, tissue} cm² g^-1)

where k contains the appropriate unit conversions (J MeV^-1, Gy kg J^-1). ⁹⁹Tc emits a 0.2935 MeV beta particle 99.998 percent of the time.

If the distance between the source and the medium of interest is long, a more precise calculation would involve considering the attenuation in the intervening air. This would require an adjustment of the beta fluence rate (f_ß).

f_ß = f_ß° e^{-(µ_{ß, air})(ρx)}

where x is the distance cm and ρ is the density of air in g cm^-3.

George Chabot, PhD