Question

a radioactive substance used in nuclear weapons decays at the rate of 6.3% per year. calculate the half - life of the radioactive substance.
the half - life of the radioactive substance is
(round to two decimal places as needed.)

Explanation:

Step1: Write the decay formula

The formula for exponential decay is \(A = A_0(1 - r)^t\), where \(A_0\) is the initial amount, \(r\) is the decay rate, and \(t\) is the time. Given \(r=0.063\). For half - life, \(A=\frac{A_0}{2}\). So, \(\frac{A_0}{2}=A_0(1 - 0.063)^t\).

Step2: Simplify the equation

Divide both sides by \(A_0\) (since \(A_0
eq0\)), we get \(\frac{1}{2}=(0.937)^t\).

Step3: Take the natural logarithm of both sides

\(\ln(\frac{1}{2})=\ln(0.937^t)\). Using the property \(\ln(a^b)=b\ln(a)\), we have \(\ln(\frac{1}{2}) = t\ln(0.937)\).

Step4: Solve for \(t\)

\(t=\frac{\ln(\frac{1}{2})}{\ln(0.937)}\). Since \(\ln(\frac{1}{2})=-\ln(2)\approx - 0.6931\) and \(\ln(0.937)\approx-0.065\), then \(t=\frac{- 0.6931}{-0.065}\approx10.66\).

Answer:

\(10.66\)