Question regarding Actvity and conversion to Gray etc

[|Fred]

Hi,

If I am not mistaken the activity of a radioactive isotope is equal to the number of atoms in the sample multiplied by the decay constant of the material. The halflife and decay constant are related by:decay constant = (ln(2))/halflife.

Would you be kind enough to check the math

If I take a sample of 3x10^10 atoms of 239Pu
ln(2)/24000 = 2.888*10^-5 yr^-1 = 9.152*10^-13 sec^-1

The activity of my sample off 3x10^10 atoms of 239Pu would be

300 * (9.152*10^-13 sec^-1) = 2.391*10^-2 atoms/sec = 0.027 Bq.

***********

Assuming this sample is ingested , this particle has an activity of 0.027Bq
So If I'm not mistaken it means that the sample will be the source 60x60x24x0.027 = 2332Bq in one day pretty much everyday if the particle is not removed.

From there how do we convert that into Gray


thank you

 

[Astronuc]

To convert disintegrations (rate) to dose (rate), one needs to know the energy of the emission, and then attenuation factors. I'll see if I can dig up some formulae.

If anyone is interested, there is a text called Radiological Health Handbook. I have the 1970 edition.

Here's something that may help. I haven't had time to review it though.

www.epa.gov/rpdweb00/docs/wipp/08-0442 attach 3.pdf

Interestingly - http://www.osti.gov/energycitations/servlets/purl/4708654-0dwoYD/4708654.pdf (use Save Target As) 25 MB.

 

[Bob S]

If I take a sample of 3x10^10 atoms of 239Pu
ln(2)/24000 = 2.888*10^-5 yr^-1 = 9.152*10^-13 sec^-1

The activity of my sample off 3x10^10 atoms of 239Pu would be

300 * (9.152*10^-13 sec^-1) = 2.391*10^-2 atoms/sec = 0.027 Bq.

***********

Assuming this sample is ingested , this particle has an activity of 0.027Bq
So If I'm not mistaken it means that the sample will be the source 60x60x24x0.027 = 2332Bq in one day pretty much everyday if the particle is not removed.

From there how do we convert that into Gray

I think we have to go back to the definition of a Gray: 1 Gray = 1 joule per kilogram, and Sieverts = Q x Grays. For the dose we should use H = DQ Sieverts, where Q is the quality factor of alphas; Q = 20. See page 10-6 in

http://www.epa.gov/oswer/riskassessment/ragsa/pdf/ch10.pdf

Then joules is alpha energy in eV times the total decays in Coulombs, because joules = volts x Coulombs. So for your number (2332 Bq) we have

J= (5.24 x 10⁶ eV) x (2332 Bq) x (1.6 x 10^-19 Coulombs/Bq) = 2 x 10^-9 joules.

So we have H = QJ/M = 20 x 2 x 10^-9 joules/M = 4 x 10^-8 joules/M Sieverts,

where M is the mass absorbing the alphas in kilograms.

===========

The range of a Pu-239 alpha is ≈ 4 cm in air = ≈ 0.004 cm in tissue.

Bob S