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A 13.83-g fragment of charcoal has been determined by carbon dating to be 4814 years old. How many decays per minute were measured in the carbon dating process? (Hint: The half-life of \({ }_{6}^{14} \mathrm{C}\) is \(5730 \mathrm{yr}\), and the \({ }_{6}^{14} \mathrm{C} /{ }_{6}^{12} \mathrm{C}\) ratio in living organic matter is \(1.20 \cdot 10^{-12} .\)

Short Answer

Expert verified
Answer: The rate of C-14 decay in the sample after 4,814 years is approximately \(2.30 \times 10^{-20}\) decays per minute.

Step by step solution

01

Find the remaining amount of C-14 after 4,814 years

We know that the half-life of C-14 is 5,730 years, and we are given that the age of the fragment is 4,814 years. First, we need to find how many half-lives have passed. To do this, we will use the formula: Number of half-lives = \(\frac{Time\ elapsed}{Half-life}\) Plugging the values, we get: Number of half-lives = \(\frac{4814\ \text{years}}{5730\ \text{years}} \approx 0.84\) half-lives Next, we need to find the remaining fraction of C-14 in the sample. To do this, we use the formula: Remaining fraction = \((1/2)^{Number\ of\ half-lives}\) Plugging the number of half-lives, we get: Remaining fraction = \((1/2)^{0.84} \approx 0.6021\)
02

Calculate the remaining C-14 in the sample

Now that we have the remaining fraction of C-14 in the sample, we can find the remaining C-14 in the sample. First, let's find the initial amount of C-14 in the sample by using the given ratio: Initial C-14 = \({}_{6}^{12} \text{C} \cdot \frac{{}_{6}^{14} \text{C}}{{}_{6}^{12} \text{C}}\) = 13.83 g \(\cdot \ 1.20 \cdot 10^{-12}\) = \(1.66 \times 10^{-11}\) g Now, we can calculate the remaining C-14 in the sample after 4,814 years: Remaining \({}_{6}^{14} \text{C}\) = \(1.66 \times 10^{-11} \text{ g} \cdot 0.6021 \approx 9.99 \times 10^{-12} \text{ g}\)
03

Calculate the decays per minute

Now that we have the remaining C-14 in the sample, we can calculate the decays per minute. First, we calculate the decay constant using the half-life of C-14: Decay constant, \(\lambda = \frac{\ln{2}}{5730\ \text{years}} \approx 1.21 \times 10^{-4}\ \text{year}^{-1}\) Then, we convert the decay constant to decays per minute: \(\lambda_{min} = 1.21 \times 10^{-4}\ \text{year}^{-1} \cdot \frac{1\ \text{year}}{525600\ \text{minutes}} \approx 2.30 \times 10^{-9} \text{ min}^{-1}\) Finally, we can calculate the decays per minute of the sample: Decays per minute = Remaining \({}_{6}^{14} \text{C} \cdot \lambda_{min}\) = \(9.99 \times 10^{-12} \text{g} \cdot 2.30 \times 10^{-9} \text{ min}^{-1} \approx 2.30 \times 10^{-20} \text{ decays/min}\) So, the rate of C-14 decay in the sample is approximately \(2.30 \times 10^{-20}\) decays per minute.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radiocarbon Decay
Radiocarbon dating, also known as carbon-14 dating, is a method used to determine the age of carbon-containing materials up to about 60,000 years old. The process is based on the radioactive decay of the carbon-14 isotope (C-14), which is found in all living organisms.

When an organism dies, it stops absorbing carbon-14 from the environment, and the C-14 it contains begins to decay at a known rate. This decay leads to a decrease in the ratio of C-14 to C-12 (the stable form of carbon) over time. By measuring the remaining amount of C-14 in a sample and comparing it to the expected C-14/C-12 ratio in a living organism, scientists can calculate the time that has passed since the organism's death.

The decay of C-14 is a random process, but statistically, it occurs at a consistent rate described by its half-life. Understanding radiocarbon decay is crucial for accurately dating archaeological finds, geological samples, and remains of historical interest.
Half-life Calculation
The half-life of a radioactive isotope is the time required for half the quantity of the substance to undergo decay. This concept is fundamental in the science of radiometric dating, including carbon dating.

Half-life is denoted by the formula: \[ T_{1/2} = \frac{\ln{2}}{\lambda} \
where \( T_{1/2} \) is the half-life and \( \lambda \) is the decay constant of the isotope. The half-life of carbon-14 is about 5,730 years, which means after this period, half of the C-14 in a sample would have decayed into nitrogen-14, its stable daughter product.

To calculate the number of half-lives that have passed, which is critical in determining the age of a sample, we divide the elapsed time since the organism's death by the half-life of C-14. Knowing the number of half-lives that have passed allows us to calculate the remaining fraction of C-14 and from there, determine the age of the sample.
C-14 Decay Constant
The decay constant (\( \lambda \)) is a parameter that characterizes the rate of radioactive decay of a substance. It is an intrinsic property of each radioactive isotope and is inversely related to the half-life. For carbon-14, the decay constant is determined by the following equation: \[ \lambda = \frac{\ln{2}}{T_{1/2}} \]In this formula, \( \ln{2} \) represents the natural logarithm of 2 (approximately equal to 0.693), and \( T_{1/2} \) is the half-life of the isotope. Since the half-life of carbon-14 is known, the decay constant can be calculated.

The knowledge of the decay constant is vital for converting the half-life information into practical measurements such as decays per minute or decays per year. This conversion is essential when determining the activity of the sample or when measuring the rate at which C-14 decays to estimate the age of archaeological and geological specimens.

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Most popular questions from this chapter

In neutron stars, which are roughly \(90 \%\) neutrons and are held together almost entirely by nuclear forces, which of the following terms become(s) relatively dominant for the binding energy, compared to that energy in an ordinary nucleus? a) the Coulomb term b) the asymmetry term c) the pairing term d) all of the above e) none of the above

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