Cobalt-60 is commonly used as a source of \(\beta\) particles. How long does it take for 87.5\(\%\) of a sample of cobalt-60 to decay (the half-life is 5.26 years)?

We are given that 87.5% of the cobalt-60 sample has decayed. This means that the remaining undecayed substance is 100% - 87.5% = 12.5%. If we represent this as a decimal, we have 0.125.

We will use the half-life formula to find the time it takes for 87.5% of the sample to decay. \[ N(t) = N_0 \cdot \left(\dfrac{1}{2}\right)^\frac{t}{T} \] Substitute N(t) with 0.125 * N_0, since 12.5% of the substance remains undecayed. \[ 0.125 N_0 = N_0 \cdot \left(\dfrac{1}{2}\right)^\frac{t}{T} \]

Divide both sides of the equation by N_0 to cancel it out. \[ 0.125 = \left(\dfrac{1}{2}\right)^\frac{t}{T} \] Now we need to solve for t. Since the equation involves an exponential function, we can use logarithms to solve for t. The easiest logarithm to use in this case is the natural logarithm (ln) as it allows us to then use the Change of Base formula. \[ \ln{0.125} = \ln{\left(\dfrac{1}{2}\right)^\frac{t}{T}} \] Using the logarithm power rule, we can bring the exponent out in front. \[ \ln{0.125} = \dfrac{t}{T} \cdot \ln{\left(\dfrac{1}{2}\right)} \] Now, divide by the \(\ln{\left(\dfrac{1}{2}\right)} \) to isolate t: \[ t = T \cdot \dfrac{\ln{0.125}}{\ln{\left(\dfrac{1}{2}\right)}} \] Substitute the half-life T with the given value 5.26 years. \[ t = 5.26 \cdot \dfrac{\ln{0.125}}{\ln{\left(\dfrac{1}{2}\right)}} \]

Now, use a calculator to calculate the value of t. \[ t \approx 5.26 \cdot \dfrac{\ln{0.125}}{\ln{\left(\dfrac{1}{2}\right)}} \approx 14.92\, years \] So, it takes approximately 14.92 years for 87.5% of a cobalt-60 sample to decay.