Question

the half-life of a radioactive kind of americium is 7,380 years. how much will be left after 22,140 years, if you start with 188,320 grams of it?
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Explanation:

Step1: Find the number of half - lives

The formula to find the number of half - lives \(n\) is \(n=\frac{t}{T}\), where \(t\) is the total time passed and \(T\) is the half - life.
Given \(t = 22140\) years and \(T=7380\) years.
So \(n=\frac{22140}{7380}=3\).

Step2: Use the half - life formula for radioactive decay

The formula for the amount of radioactive substance left \(A\) after \(n\) half - lives is \(A = A_0\times(\frac{1}{2})^n\), where \(A_0\) is the initial amount.
Given \(A_0 = 188320\) grams and \(n = 3\).
Substitute the values into the formula: \(A=188320\times(\frac{1}{2})^3\).
First, calculate \((\frac{1}{2})^3=\frac{1}{8}\).
Then, \(A = 188320\times\frac{1}{8}\).
\(188320\div8 = 23540\).

Answer:

23540