How Do You Calculate the Effective Dose from a Given Concentration?

[Smed]

How do you get from a concentration to an effective dose? For example, if you have 10^-9 Ci/cm³ of tritium in a human body, then how do you get the effective dose to that body?

Thanks

 

[HotCells]

Using the Mass Activity Calculator in Nucleonica (www.nucleonica.net), 10^-9 Ci of tritum corresponds to 37 Bq. Also from Nucleonica, the effective dose coefficient for ingestion for tritium is 1.8x10^-11 Sv/Bq. Hence the dose is 37 x 1.8x10^-11 Sv per cm³ = 57 x 10^-11 Sv per cm³. To obtain the total dose you then need to multiply by the volume of the human body or that part which has been exposed- I'll leave this as an exercise! If the tritium intake is by inhalation, then the effective dose coefficient for inhalation must be used. This is 2.6x10^-10 Sv per Bq i.e. considerably higher than that for ingestion.

 

[Smed]

I'm not sure I follow how time plays into this. Does this process give me the Sv per second or per disintegration? If you have a decaying source, you wouldn't just get a single dose. I guess what I'm looking for is the dose rate, sorry for the confusion.

 

[Astronuc]

How do you get from a concentration to an effective dose? For example, if you have 10^-9 Ci/cm³ of tritium in a human body, then how do you get the effective dose to that body?

Thanks

One could integrate does rate per unit volume over time and get a does per unit volume, which of course assumes that all that is produced is absorbed. Of course, one would take into account the decay of the nuclide as well as the biological removal rate.

 

[HotCells]

I'm not sure I follow how time plays into this. Does this process give me the Sv per second or per disintegration? If you have a decaying source, you wouldn't just get a single dose. I guess what I'm looking for is the dose rate, sorry for the confusion.

Good question. Actually the effective dose coefficients are denoted by e(50). The 50 refers to the following 50 years after incorporation. Sophisticated bio-kinetic models have been developed to account for the total radiation damage to the cells from the incorporation of a radionuclide. All known effects are accounted for e.g. biological half-life, presence of daughters etc. The relationship between the Dose and the Activity is given by

Dose (Sv) = e(50) x Activity (Bq)

From this value one can estimate the risk of cancer for example. Basically the unit Sievert (Sv) is the energy deposited in the tissue.

If you would like to calculate the dose rate this requires a different approach. From the activity in Bq, one obtains the number of disintegrations per second. In each disintegration, a 20 keV electron is emitted. This electron is then absorbed in the tissue causing damage. So the rate of energy released is

Energy emission rate (keV/s) = Activity (Bq) x 20 keV

with units keV per second. This needs to be converted into joules i.e.

Energy emission rate (Joule per second) = activity x 20 x 1000 x 1.6e-19

For the activity of 37 Bq, this corresponds to

Energy emission rate (joule per second)= 37 x 20 x1000 x 1.6e-19 = 1.2 E-16

Since the "quality factor" for electrons is one, this means that the dose rate is equal to the energy emission rate i.e.
Dose Rate = 1.2 E- 16 Sieverts per second.

This is an extremely small dose rate.

For further information, I refer you to the relevant Nucleonica wiki article...
http://www.nucleonica.net/wiki/index.php/Help:Dosimetry_%26_Shielding