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The practical limit to ages that can be determined by radio carbon dating is about 41 000 yr. In a 41 000-yr-old sample, what percentage of the original \({ }_{6}^{14} \mathrm{C}\) atoms remains?

Short Answer

Expert verified
0.743% of the original \( {}_{6}^{14} \mathrm{C} \) remains.

Step by step solution

01

Understand Radioactive Decay

The decay of radioactive substances is described by the exponential decay model. For carbon-14, we use the half-life formula to calculate the remaining percentage of the substance.
02

Determine the Half-Life of Carbon-14

The half-life of carbon-14 is approximately 5730 years. This means every 5730 years, half of the original carbon-14 has decayed.
03

Calculate Number of Half-Lives

To find out how many half-lives have passed in 41,000 years, divide the total age by the half-life of carbon-14: \( n = \frac{41000}{5730} \approx 7.157 \) half-lives.
04

Calculate Remaining Carbon-14

Use the formula for exponential decay: \( N = N_0 \times \left(\frac{1}{2}\right)^n \), where \( n \) is the number of half-lives. Here, \( N_0 \) is the initial quantity (100%), and \( N \) is the remaining quantity.
05

Calculate Percentage of Remaining Atoms

Substitute \( n = 7.157 \) into the equation: \( N = 100 \times \left(\frac{1}{2}\right)^{7.157} \approx 0.743\% \).
06

Conclusion

After decaying for 41,000 years, approximately 0.743% of the original carbon-14 atoms remain in the sample.

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Key Concepts

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

Half-Life
The concept of half-life is pivotal to understanding radioactive decay. Half-life is defined as the amount of time it takes for half of the radioactive nuclei in a sample to decay. This means that after one half-life, 50% of the initial radioactive atoms have transformed into another substance.
For example, if you start with 100 atoms of a radioactive substance and the half-life is set, after one half-life, you will have 50 atoms remaining. After another half-life, only 25 atoms will be left.

Half-life is important because it helps scientists determine the age of objects containing radioactive materials. By measuring a sample and knowing the half-life of the substance within it, they can estimate how long the substance has been decaying and thus date the object. For carbon-14, which is vital in archaeological studies, the half-life is 5730 years.
Carbon-14 Dating
Carbon-14 dating, often referred to as radiocarbon dating, is a method used to determine the age of an object containing organic material. This is done by evaluating the amount of carbon-14 remaining in that object.
Here's how it works:
  • Carbon-14 is a radioactive isotope of carbon that is found in all living organisms.
  • When an organism dies, it stops absorbing carbon-14; the isotopes then begin to decay gradually.
  • By calculating how much carbon-14 remains in the sample, and knowing the half-life of carbon-14, scientists can work backward to figure out when the organism died, thereby dating the specimen.
The effectiveness of carbon-14 dating diminishes beyond about 50,000 years. This is because after so many years, the fraction of carbon-14 left in a sample is so small that it becomes difficult to measure accurately. For the original problem, this is why it's only practical to determine ages up to around 41,000 years, as suggested.
Exponential Decay
Radioactive decay follows an exponential decay model, which means it reduces by a constant percentage over regular intervals of time. In the case of carbon-14, this interval is its half-life of 5730 years.
Mathematically, exponential decay can be described using the formula:
\[ N = N_0 \times \left(\frac{1}{2}\right)^n \] Here, \( N \) is the amount remaining after \( n \) half-lives, and \( N_0 \) is the initial amount. As each half-life passes, the number of radioactive atoms decreases by half, leading to an exponential curve.

This concept is crucial for radioactive dating methods. With the formula, scientists can plug in known values (such as the half-life and elapsed time since death) to calculate the remaining atoms present in a sample. From there, they can determine the "age" of the sample and, therefore, its timeline in history.

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Most popular questions from this chapter

The half-life for the \(\alpha\) decay of uranium \(\frac{238}{92} \mathrm{U}\) is \(4.47 \times 10^{9} \mathrm{yr} .\) Determine the age (in years) of a rock specimen that contains sixty percent of its original number of \(\frac{238}{92} \mathrm{U}\) atoms.

To make the dial of a watch glow in the dark, \(1.000 \times 10^{-9} \mathrm{~kg}\) of radium \(\frac{226}{88} \mathrm{Ra}\) is used. The half-life of this isotope has a value of \(1.60 \times 10^{3}\) yr. How many kilograms of radium disappear while the watch is in use for fifty years?

Osmium \({ }_{76}^{191} \mathrm{Os}\) (atomic mass \(=190.960920 \mathrm{u}\) ) is converted into Iridium \(\frac{191}{77} \mathrm{Ir}\) (atomic mass \(=190.960584 \mathrm{u}\) ) via \(\beta^{-}\) decay. What is the energy (in \(\mathrm{MeV}\) ) released in this process?

Osmium \(\frac{191}{76} \mathrm{Os}\) (atomic mass \(\left.=190.960920 \mathrm{u}\right)\) is converted into Iridium \(\frac{191}{77} \mathrm{Ir}\) (atomic mass \(=190.960584 \mathrm{u}\) ) via \(\beta^{-}\) decay. What is the energy (in \(\mathrm{MeV}\) ) released in this process?

(a) Energy is required to separate a nucleus into its constituent nucleons, as Figure \(31-3\) indicates; this energy is the total binding energy of the nucleus. In a similar way one can speak of the energy that binds a single nucleon to the remainder of the nucleus. For example, separating nitrogen \({ }_{7}^{14} N\) into nitrogen \({ }_{7}^{13} \mathrm{~N}\) and a neutron takes energy equal to the binding energy of the neutron, as shown below: $$ \frac{14}{7} N+\text { Energy } \rightarrow \frac{13}{7} N+\frac{1}{0} n $$ Find the energy (in MeV) that binds the neutron to the \(\frac{14}{7} N\) nucleus by considering the mass of \(13 \mathrm{~N}\) (atomic mass \(=13.005738 \mathrm{u}\) ) and the mass of \(\frac{1}{0^{\mathrm{n}}}\) (atomic mass \(=1.008\) \(665 \mathrm{u}\) ), as compared to the mass of \({ }_{7}^{14} \mathrm{~N}\) ( atomic mass \(=14.003074 \mathrm{u}\) ). (b) Similarly, one can speak of the energy that binds a single proton to the\(\frac{14}{7} \mathrm{~N}\) nucleus: $$ { }_{7}^{14} \mathrm{~N}+\text { Energy } \rightarrow{ }_{6}^{13} \mathrm{C}+{ }_{1}^{1} \mathrm{H} $$ Following the procedure outlined in part (a), determine the energy (in MeV) that binds the proton (atomic mass \(=1.007825 \mathrm{u}\) ) to the \(\frac{14}{7} \mathrm{~N}\) nucleus. The atomic mass of carbon \(\frac{13}{6} \mathrm{C}\) is \(13.003335 \mathrm{u}\) (c) Which nucleon is more tightly bound, the neutron or the proton?

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