SOLUTION: An isotope has a​ half-life of 92 years. How much of a 27​-gram sample is left after 150 ​years?

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Question 1104707: An isotope has a​ half-life of 92 years. How much of a 27​-gram sample is left after 150 ​years?
Found 3 solutions by josgarithmetic, Boreal, greenestamps:
Answer by josgarithmetic(39620) About Me  (Show Source):
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If your model is to be y=pe%5E%28-kt%29, then k=ln%282%29%2F92, or k=0.007534.

The question makes for y=27%2Ae%5E%28-0.007534%2A150%29.

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
P=Poe-kt
(1/2)=e^-kt
ln(1/2)=-kt
-k=ln(0.5)/92=-0.0075
k=0.0075
P=27*e^(-0.0075*150)=27*e^(-1.125)=27*0.3247=8.77 gm.

Answer by greenestamps(13200) About Me  (Show Source):
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If you are going to use the Ae%5E%28-kt%29 formula, and you want a high degree of accuracy in your answer, then you need to keep several digits of the logarithm for your calculations.

One of the tutors who answered your question got an answer of 8.77 grams; a more accurate answer is 8.72 grams. The better accuracy requires a more accurate value for the logarithm involved in the calculation.

But you can avoid the logarithm problem completely by using a purely mathematical calculation, instead of using the Ae%5E%28-kt%29 formula.

The amount of the original sample after n half-lives is the original amount, multiplied by one-half raised to the power n.

In this problem, the number of half-lives is 150/92. So the amount remaining after 150 years is
27%28.5%29%5E%28150%2F92%29+=+8.72075 to 5 decimal places.

There is no need with this calculation to worry about rounding errors....