- Identify the decay formula and set the remaining amount \(N(t)\) to \(10\%\) of the initial amount \(N_0\).
- Simplify the resulting equation \(0.10 = e^{-\frac{t \ln 2}{h}}\) by taking the natural logarithm of both sides.
- Solve for \(t\) by isolating the variable, resulting in the expression \(t = h \cdot \frac{\ln 10}{\ln 2}\).
- Substitute the given half-life \(h = 27.7\) days to identify the final expression: \(\boxed{t = (27.7) \left( \frac{\ln 10}{\ln 2} \right)}\).
Explanation
- Analysis of the Radioactive Decay Problem
The radioactive decay of a substance is modeled by the formula \(N(t) = N_0 e^{-\frac{t \ln 2}{h}}\), where \(N_0\) is the initial amount of the substance, \(h\) is its half-life, and \(N(t)\) is the amount remaining at time \(t\). We are given that the initial amount of Chromium-51 is \(N_0 = 85\) milligrams, and its half-life is \(h = 27.7\) days. We want to find the time \(t\) required for \(90\%\) of the material to decay. If \(90\%\) has decayed, the amount remaining is \(10\%\) of the initial amount, i.e., \(N(t) = 0.10 N_0\).
- Solving for Time t
To find \(t\), we substitute \(N(t) = 0.10 N_0\) into the decay formula:
\[0.10 N_0 = N_0 e^{-\frac{t \ln 2}{h}}\]
Dividing both sides by \(N_0\) (which is \(85\) mg, thus non-zero), we get:
\[0.10 = e^{-\frac{t \ln 2}{h}}\]
To isolate \(t\), we take the natural logarithm (\(\ln\)) of both sides:
\[\ln(0.10) = -\frac{t \ln 2}{h}\]
Since \(0.10 = \frac{1}{10}\), we can write this as:
\[\ln\left(\frac{1}{10}\right) = -\frac{t \ln 2}{h}\]
Using the property of logarithms \(\ln(\frac{1}{x}) = -\ln(x)\), we know \(\ln(\frac{1}{10}) = -\ln(10)\). Thus:
\[-\ln(10) = -\frac{t \ln 2}{h}\]
Multiplying both sides by \(-1\):
\[\ln(10) = \frac{t \ln 2}{h}\]
Solving for \(t\):
\[t = h \cdot \frac{\ln(10)}{\ln(2)}\]
Substituting the value for the half-life \(h = 27.7\):
\[t = 27.7 \cdot \frac{\ln(10)}{\ln(2)}\]
- Conclusion and Final Answer Selection
Comparing our result \(t = 27.7 \cdot \frac{\ln 10}{\ln 2}\) with the given options, we find it matches the third option exactly. Therefore, the time it takes for 90% of the material to decay is \(\boxed{t = (27.7) \left( \frac{\ln 10}{\ln 2} \right)}\).
Examples
Radioactive decay is a fundamental concept used in medicine for diagnostic purposes. For example, Chromium-51 is used to label red blood cells to determine their lifespan or to diagnose certain blood disorders. By understanding how the substance decays over time, doctors can calculate the precise dose required for a test, ensuring the radioactivity remains at safe levels while still being detectable by scanners. This mathematical modeling allows for accurate medical procedures, helping patients receive effective treatments while minimizing radiation exposure.