Question #120074
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.
1
Expert's answer
2020-06-04T09:59:31-0400

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


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

where A(0)=1900 MBqA(0) = 1900\space MBq is the activity at the initial moment of time and T=6 hoursT = 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 MBq2256=105.8 MBqA(25) = 1900\space MBq \cdot 2^{-\frac{25}{6}} = 105.8\space MBq

To convert this number to Ci, one should multiply it by 2.710112.7\cdot 10^{-11} . Thus:


A=105.8106 Bq=105.81062.71011Ci2.86 mCiA = 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.


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