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? grams

Explanation:

Step1: Find the number of half - lives

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

Step2: Use the radioactive decay formula

The radioactive decay formula is \(N = N_0\times(\frac{1}{2})^n\), where \(N_0\) is the initial amount, \(n\) is the number of half - lives, and \(N\) is the final amount.
We know that \(N_0 = 188320\) grams and \(n = 3\).
First, calculate \((\frac{1}{2})^3=\frac{1}{8}\)
Then, \(N=188320\times\frac{1}{8}\)
\(188320\div8 = 23540\)

Answer:

23540