Radiocarbon dating, or carbon-14 dating, is a scientific method that can accurately determine the age of organic materials as old as approximately 60,000 years. The technique is based on the decay of the carbon-14 isotope. The time in which half of the original number of atom decay is defined as the half-life. It is modelled by the equation N=N0e−ktN = N_0 e^{-kt}, where kk is a constant and tt is the time in years. N0N_0 is the initial amount of material in an object. Carbon-14 has a half life of 5730 years, meaning that 5730 years after an organism dies, half of its carbon-14 atoms have decayed.

  1. (a)

    Find the value of kk.

  2. (b)

    A mummified body was found to have 8% of its original atoms left. How many years ago did the person die? Leave your answer to the nearest whole number.

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Worked solution

3(a)

1. Substitute the half-life conditions into the given model

When t=5730t = 5730, N=12N0N = \frac{1}{2}N_0.
12N0=N0e−5730k\frac{1}{2}N_0 = N_0 e^{-5730k}

2. Solve for $k$ using natural logarithms

0.5=e−5730k0.5 = e^{-5730k}
ln⁡0.5=−5730k\ln 0.5 = -5730k
k=ln⁡0.5−5730k = \frac{\ln 0.5}{-5730}
k≈0.000120968≈0.000121 (to 3 s.f.)k \approx 0.000120968 \approx 0.000121 \text{ (to 3 s.f.)}

Answer

0.0001210.000121

3(b)

1. Substitute the given percentage and the value of $k$ from part (a) into the model

From part (a), k=ln⁡25730k = \frac{\ln 2}{5730}.
Given N=0.08N0N = 0.08N_0,
0.08N0=N0e−(ln⁡25730)t0.08N_0 = N_0 e^{-\left(\frac{\ln 2}{5730}\right)t}

2. Solve for $t$ using logarithms

0.08=e−(ln⁡25730)t0.08 = e^{-\left(\frac{\ln 2}{5730}\right)t}
ln⁡0.08=−(ln⁡25730)t\ln 0.08 = -\left(\frac{\ln 2}{5730}\right)t
t=−5730ln⁡0.08ln⁡2t = -\frac{5730 \ln 0.08}{\ln 2}
t≈20879.43t \approx 20879.43

3. State the final answer rounded to the nearest whole number

t=20879 (to nearest whole number)t = 20879 \text{ (to nearest whole number)}

Answer

2087920879 years

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