The half-life of Cobalt-60 is 5.27 years.
The radioactive decay of Cobalt-60 follows an exponential decay model given by the formula \(N(t) = N_0 e^{-kt}\), where \(N(t)\) is the quantity remaining at time \(t\), \(N_0\) is the initial quantity, and \(k\) is the decay rate constant. We are given the half-life \(T_{1/2} = 5.27\) years and the initial amount \(N_0 = 1000\) mg. Our goal is to find the decay constant \(k\) and the remaining amount after \(t = 7\) years.
The decay constant \(k\) is related to the half-life by the equation \(T_{1/2} = \frac{\ln(2)}{k}\). Rearranging this formula to solve for \(k\), we get:
\[k = \frac{\ln(2)}{T_{1/2}}\]
Substituting \(T_{1/2} = 5.27\):
\[k = \frac{\ln(2)}{5.27} \approx \frac{0.69314718}{5.27} \approx 0.131527\]
Rounding to six decimal places, we find the decay rate constant \(k = 0.131527\).
To find the amount remaining after \(t = 7\) years, we use the formula \(N(t) = N_0 e^{-kt}\). Substituting \(N_0 = 1000\), \(k = 0.131527\), and \(t = 7\):
\[N(7) = 1000 \cdot e^{-0.131527 \cdot 7}\]
Performing the calculation:
\[N(7) = 1000 \cdot e^{-0.920689} \approx 1000 \cdot 0.3982446 \approx 398.2446\]
Rounding the result to two decimal places, we find that the amount remaining after 7 years is approximately \(398.24\) mg.
The decay rate of Cobalt-60 is \(0.131527\) per year, and the amount remaining after 7 years from a 1000 mg sample is \(\boxed{398.24}\) mg.
Radioactive decay is vital in fields like medicine and archaeology. For instance, Cobalt-60 is used in radiotherapy to treat cancer; doctors must calculate its precise decay rate to determine the exact dosage delivered to a patient over time as the source strength changes. Similarly, in fields like radiocarbon dating, scientists use the known decay rate of isotopes to estimate the age of ancient artifacts, providing a scientific window into the past.