Question

if 100 g of au - 198 decays to 6.25 g in 10.8 days, how many days is the half - life of au - 198?

Explanation:

Step1: Determine the number of half - lives

The formula for radioactive decay is \(N = N_0\times(\frac{1}{2})^n\), where \(N\) is the final amount, \(N_0\) is the initial amount, and \(n\) is the number of half - lives.
Given \(N_0=100\space g\), \(N = 6.25\space g\).
Substitute into the formula: \(6.25=100\times(\frac{1}{2})^n\)
\(\frac{6.25}{100}=(\frac{1}{2})^n\)
\(0.0625 = (\frac{1}{2})^n\)
Since \(\frac{1}{16}=(\frac{1}{2})^4\), so \(n = 4\)

Step2: Calculate the half - life

We know that the total time \(t = 10.8\) days and the number of half - lives \(n = 4\)
The formula for the half - life \(T=\frac{t}{n}\)
Substitute \(t = 10.8\) days and \(n = 4\) into the formula: \(T=\frac{10.8}{4}=2.7\) days

Answer:

The half - life of Au - 198 is \(2.7\) days.