Asked by shannon ortiz
iodine 131 has a half life of 8 days. If 5 mg are stored for a week, how much is left? How many days does it take before only 1 mg remains?
An explanation for each step would be nice please. I want to learn how to do these problems for my upcoming exam.
An explanation for each step would be nice please. I want to learn how to do these problems for my upcoming exam.
Answers
Answered by
Steve
every time 8 days pass, you multiply by 1/2. So, after t days, 8 days have passed t/8 times, and the amount left is
5*(1/2)^(t/8)
After 7 days, then, you have
5*0.5^(7/8) = 2.726mg
1mg is 1/5 of the starting amount. 1/5 is a bit less than 1/4 = (1/2)^2, so it should take a little more than 2 half-lives, or 16 days.
5(0.5)^(t/8) = 1
0.5^(t/8) = 0.2
t/8 = log0.2/log0.5
t = 8 log0.2/log0.5 = 18.575 days
5*(1/2)^(t/8)
After 7 days, then, you have
5*0.5^(7/8) = 2.726mg
1mg is 1/5 of the starting amount. 1/5 is a bit less than 1/4 = (1/2)^2, so it should take a little more than 2 half-lives, or 16 days.
5(0.5)^(t/8) = 1
0.5^(t/8) = 0.2
t/8 = log0.2/log0.5
t = 8 log0.2/log0.5 = 18.575 days
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