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The amount of a sample remaining after $t$ days is given by the equation $P(t)=A\left(\frac{1}{2}\right)^{\frac{t}{h}}$, where $A$ is the initial amount of the sample and $h$ is the half-life, in days, of the substance. A sample contains $60\%$ of its original amount of Fermium-257. The half-life of Fermium-257 is about 100 days. About how old is the sample?


A60 days
B74 days
C136 days

Verified Answer :

  • Set up the decay equation \(0.60A = A(0.5)^{t/100}\).
  • Simplify and apply logarithms to isolate the time variable \(t\).
  • Calculate \(t = 100 \cdot \frac{\ln(0.6)}{\ln(0.5)} \approx 73.7\) days.
  • Conclude that the sample is approximately \(\boxed{74}\) days old, matching option (a).

Explanation

  1. Analyzing the Decay Formula

The radioactive decay of a substance is governed by the formula \(P(t) = A\left(\frac{1}{2}\right)^{\frac{t}{h}}\), where \(P(t)\) is the amount remaining at time \(t\), \(A\) is the initial amount, and \(h\) is the half-life of the substance. For this problem, we are given the half-life of Fermium-257 as \(h = 100\) days. We are also told that the sample currently contains \(60\%\) of its original amount, meaning \(P(t) = 0.60A\). Our objective is to determine the age of the sample, \(t\), in days.

  1. Setting Up the Equation for t

Substitute the given values into the radioactive decay equation:
\[0.60A = A\left(\frac{1}{2}\right)^{\frac{t}{100}}\]
Dividing both sides by the initial amount \(A\), we obtain the equation:
\[0.60 = \left(0.5\right)^{\frac{t}{100}}\]
To solve for \(t\), we take the natural logarithm (\(\ln\)) of both sides:
\[\ln(0.60) = \ln\left(0.5^{\frac{t}{100}}\right)\]
Using the logarithmic power property \(\ln(x^y) = y \cdot \ln(x)\), this becomes:
\[\ln(0.60) = \frac{t}{100} \cdot \ln(0.5)\]

  1. Solving for Time t

We isolate \(t\) by multiplying both sides by \(100\) and dividing by \(\ln(0.5)\):
\[t = 100 \cdot \frac{\ln(0.60)}{\ln(0.5)}\]
Using the values \(\ln(0.60) \approx -0.5108\) and \(\ln(0.5) \approx -0.6931\), we calculate:
\[t \approx 100 \cdot \frac{-0.5108}{-0.6931} \approx 100 \cdot 0.7370 \approx 73.70\]
Rounding to the nearest whole number, we find that the sample is approximately \(74\) days old. This corresponds to option (a).

Examples

Radioactive decay, as demonstrated by the formula used here, is a fundamental concept in fields like archaeology and medicine. For example, archaeologists use carbon-14 dating to determine the age of ancient organic remains. By knowing the half-life of carbon-14 and measuring the amount remaining in a sample, scientists can calculate how many thousands of years have passed since the organism died, providing an essential timeline for historical and prehistoric discoveries.

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