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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?

 Dec 8, 2021
 #1
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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.

 Dec 8, 2021
 #2
avatar+2287 
+1

Here’s a faster solution method:

 

\(\large \text{half-life} = \large \dfrac{29} {Log_2(6272) – Log_2(196)} = 5.8 \text { minutes} \)

 

 

 

GA

--. .-

 Dec 10, 2021

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