Question 1190552
Cesium 137 is a radioactive metal with a short half-life of 30 years. In a
sample originally having 1 gram of cesium 137, the amount of cesium 137 present after t years is given by 
A(t) = (1/2)^t/30. where A is the amount of cesium in grams and t is the time in years.
a. How much cesium 137 will be present after 30 years?
b. How much cesium 137 will be present after 90 years?
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<pre>
(a)  30 years is one half-life.


     THEREFORE, the remaining mass of the 1 gram of cesium-137 after 30 years is  {{{1/2}}} = 0.5 grams.


     You will get the same answer, if you substitute the value t= 30 years into the given exponential decay function.





(b)  90 years is 3 (three) half-lives.


     THEREFORE, the remaining mass of the 1 gram of cesium-137 after 90 years is  {{{(1/2)^3}}} = {{{1/2^3}}} = {{{1/8}}} = 0.125 grams.


     You will get the same answer, if you substitute the value t= 90 years into the given exponential decay function.

</pre>

Solved.


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On radioactive decay, &nbsp;see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/logarithm/Radioactive-decay-problems.lesson>Radioactive decay problems</A> 

in this site.


You will find many similar &nbsp;(and different) &nbsp;solved problems there.



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&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic "<U>Logarithms</U>".



Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


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