Answer to Question #120074 in Atomic and Nuclear Physics for Raghuveer Reddy

Question

2. A vial contains 1900 MBq of Technetium-99m on the 18th June 2014 at 06:00. How much will it contain at 07:00 on the 19th June 2014? Please input your answer in Bq and Ci.

Answer 1

According to the radiactive decay law, the activity of the sample at the time t is given by the expression:

$$A(t) = A(0) \cdot 2^{-\frac{t}{T}}$$

where $A(0) = 1900\space MBq$ is the activity at the initial moment of time and $T = 6\space hours$ is the half-life of Technetium-99m.

The time was passed between 06:00, 18th June 2014 and 07:00, 19th June 2014 is 25 hours. Thus, the activity will be:

$$A(25) = 1900\space MBq \cdot 2^{-\frac{25}{6}} = 105.8\space MBq$$

To convert this number to Ci, one should multiply it by $2.7\cdot 10^{-11}$ . Thus:

$$A = 105.8\cdot 10^{6}\space Bq = 105.8\cdot 10^{6}\cdot 2.7\cdot 10^{-11} Ci \approx 2.86\space mCi$$

Answer. A = 105.8*10^6, A = 2.86*10^(-3) Ci.