What is the half-life of a radioactive isotope if the activity of a sample drops from 6,272 cpm to 196 cpm in 29.0 min?
196 ==6,272 x (1/2)^(29/t), solve for t
196 /6,272 ==(1/2)^(29/t)
0.03125 ==1/2^(29/t)
Take the log of both sides
29/t ==log(0.03125) / log(1/2)
29/t ==5
5t ==29
t ==29 /5 ==5.8 minutes - half-life of a radioactive isotope.
Here’s a faster solution method:
\(\large \text{half-life} = \large \dfrac{29} {Log_2(6272) – Log_2(196)} = 5.8 \text { minutes} \)
GA
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