Question

if the half - life of a radioactive substance is 5,800 years, then at what rate is it decaying? (hint: the amount goes from $a_0$ to $\frac{1}{2}a_0$ in 5,800 years.) the decay rate, k, is. (round to six decimal places as needed.)

Explanation:

Step1: Write the exponential decay formula

The exponential decay formula is \( A = A_0e^{kt}\), where \(A\) is the amount of the substance at time \(t\), \(A_0\) is the initial amount, \(k\) is the decay rate, and \(t\) is the time.

Step2: Substitute the given values into the formula

We know that when \(t = 5800\), \(A=\frac{1}{2}A_0\). Substituting these into the formula \(\frac{1}{2}A_0=A_0e^{k\times5800}\).
Since \(A_0
eq0\), we can divide both sides of the equation by \(A_0\), getting \(\frac{1}{2}=e^{5800k}\).

Step3: Take the natural logarithm of both sides

Taking the natural logarithm of both sides: \(\ln(\frac{1}{2})=\ln(e^{5800k})\).
Using the property \(\ln(e^{x}) = x\), the right - hand side simplifies to \(5800k\). And \(\ln(\frac{1}{2})=-\ln(2)\). So, \(-\ln(2)=5800k\).

Step4: Solve for \(k\)

Solving for \(k\), we have \(k =-\frac{\ln(2)}{5800}\).
Calculating \(\ln(2)\approx0.693147\), then \(k=-\frac{0.693147}{5800}\approx - 0.00011951\).

Answer:

\(-0.000120\)