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Answer to Question #295 Submitted to "Ask the Experts"Category: Radiation Basics The following question was answered by an expert in the appropriate field: Q
Given a known amount of a radioactive compound, say 1 oz. thorium nitrate, how would you calculate the aproximate activity of that amount in Ci.?
A
To do this you need to (1) determine the important molecular, atomic, isotopic, and radioactivity factors about the material, (2) set up the equation for the specific activity of the particular radionuclide(s), (3) determine the mass of the particular radionuclide(s), and (4) calculate the activity.
1. Facts about thorium nitrate and thorium
Thorium nitrate: Th(NO3)4, Molecular weight = 480.06.
Thorium: Natural thorium is 100 percent 232Th; Atomic Weight (A) = 232.08;
half-life (T1/2) =1.40 × 1010 years.
2. Equation for Specific Activity of 232Th
(Note: This part is simplified because natural thorium has only one isotope. If this question were about a compound of uranium, the calculation would have to consider the fractional contributions of each of the three radioactive isotopes in natural uranium.)
Specific Activity for any radionuclide with decay constant k:
Recall that N, atoms/g =(6.023 × 1023 atoms/mole)/A, g/mole. Then:
SA (Bq/g) = kN = (0.693/T1/2) × (6.023 × 1023)/A
= (4.174 × 1023)/(T1/2 × A) for T1/2 in seconds; or
= (1.323 × 1016)/(T1/2 × A) for T1/2 in years.
In Ci, where 1 Ci = 3.7 × 1010 Bq (or 3.7 × 1010 transformations/s):
SA (Ci/g) =(SA, Bq/g)/(3.7 × 1010 Bq/Ci)
= (1.128 × 1013)/(T1/2 × A) for T1/2 in seconds; or
= (3.575 × 105)/(T1/2 × A) for T1/2 in years.
Specific Activity for 232Th:
SA = (1.323 × 1016)/(1.40 × 1010 × 232.038) = 4.07 × 103 Bq/g, or
(4.072 × 103 Bq/g)/(3.7 × 1010 Bq/Ci) = 1.10 × 10-7 Ci/g.
3. Mass of Thorium:
Th(NO3)4: 1 oz. × 28.35 g/oz = 28.35 g.
Th: 28.35 g × 232.038/480.06 = 13.70 g.
4. Activity of 1 oz (28.35 g) of Thorium Nitrate containing 13.70 g
of 232Th:
Activity = 13.7 g × 4.07 × 103 Bq/g = 5.58 × 104 Bq or 55.8 kBq or
13.7 g × 1.10 × 10-7 Ci/g = 1.51 × 10-6 Ci or 1.51 µCi.
Note: This was for anhydrous Th(NO3)4; appropriate adjustments would have to be made if the Th(NO3)4 contains water of hydration.
C. E. Roessler, Ph.D., CHP
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