First you have to determine how many half-lives have past for the iodine sample. Keep in mind that with each passing half-life, the total amount of I-126 remaining is decreased by 1/2. Therefore, the total amount of I-126 remaining after a certain number of half-lifes is:
A = Ao x (1/2)^n
where A is the current amount, Ao is the original amount, and n is the number of half-lifes.
So let's do some plug'n'chug:
(3.29 g) = (26.3 g)(1/2)^n
0.125 = (1/2)^n
We need to use logs to solve for n:
log(0.125) = log((1/2)^n)
log(0.125) = n*log(1/2)
-0.903 = (-0.301)n
n = 3 half-lives.
Since the length of 1 half-life is 13 days, you can then say that [3 x 13 days =] 39 days have passed.
2007-09-23 08:26:26
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answer #1
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answered by Lucas C 7
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Based on the rule that the half life of a radioactive substance is the time it takes for half of the sample to degrade I have calculated the following: 26.32 / 2 = 13.16 > 13.16 / 2 = 6.58> 6.58 / 2 = 3.29 and based on the accepted half life of Iodine-126 : 13 Days, the sample is approximately 39 days old give or take a few hours depending on the exact mass of the sample when measured to more significant figures.
2007-09-23 08:43:08
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answer #2
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answered by Johnathan C 1
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26.3 /2 = 13.15 /2 = 6.575 /2 =3.29 3 half lives were used.
13*3 = 39 days
2007-09-23 08:21:44
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answer #3
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answered by Kyle R 2
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Every 13 days, the amount remaining will be divided in half. After the first 13 days, there would be 13.15g left. After another 13 days, you have 6.58g. One more half-life gives you your answer - 3.29g.
This means 39 days have passed, and you have divided in half 3 times.
2007-09-23 08:25:05
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answer #4
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answered by skeptik 7
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log(26.3/3.29) = log2^x
log(26.3/3.29) = x log 2
log7.99392 = x log2
0.90275 = x 0.30103
x = 0.90275/0.30103
x = 2.9982 (Approx 3)
age = 3 x 13 = 39 days.
2007-09-23 08:31:40
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answer #5
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answered by lenpol7 7
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13 x 3 days, since the mass has decreased to 1/8 of its value - 3 half lives.
2007-09-23 08:22:12
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answer #6
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answered by Gervald F 7
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39 days
26.32/2=13.16
13.26/2=6.58
6.58/2=3.29
2007-09-23 08:21:37
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answer #7
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answered by Sherlock 2
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