/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 A bone fragment found in a cave ... [FREE SOLUTION] | 91Ó°ÊÓ

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A bone fragment found in a cave believed to have been inhabited by early humans contains \(0.29\) times as much \({ }^{14} \mathrm{C}\) as an equal amount of carbon in the atmosphere when the organism containing the bone died. (See Example \(19.4\) in Section 19.4.) Find the approximate age of the fragment. \((\ln 0.29=-1.209)\)

Short Answer

Expert verified
The bone fragment is approximately 9991 years old.

Step by step solution

01

Understand the Decay Formula

The age of the bone fragment can be determined using the radioactive decay formula for Carbon-14, which is \( N(t) = N_0 \times e^{-kt} \). Here, \( N(t) \) is the amount of Carbon-14 currently in the sample, \( N_0 \) is the initial amount when the organism died, \( k \) is the decay constant, and \( t \) is the time elapsed since the organism died.
02

Use the Decay Percentage

The problem states that the bone contains \(0.29\) times the original amount of Carbon-14. Using the formula \( \frac{N(t)}{N_0} = 0.29 = e^{-kt} \), we take the natural logarithm of both sides to solve for time \( t \). This results in \( \ln(0.29) = -kt \).
03

Apply Natural Logarithm Information

We're given that \( \ln(0.29) = -1.209 \), so we can substitute this value into the equation and write \( -1.209 = -kt \). This simplifies to \( kt = 1.209 \).
04

Calculate the Decay Constant

The decay constant \( k \) for Carbon-14 is known (from radiocarbon dating resources) to be \( k = \frac{\ln(2)}{5730} \) years\(^{-1}\), since the half-life of Carbon-14 is about \(5730\) years.\( \ln(2) \approx 0.693 \).
05

Solve for the Age of the Fragment

Using \( k = \frac{0.693}{5730} \approx 0.000121 \), substitute this into \( kt = 1.209 \) to find \( t \). \( t = \frac{1.209}{0.000121} \approx 9991 \) years.

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Key Concepts

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

Radioactive Decay
Radioactive decay is a natural process in which unstable atomic nuclei lose energy by emitting radiation. This process results in the transformation of an element into a different isotope or a different element altogether. In the context of Carbon-14 dating, radioactive decay is vital as it helps determine the age of organic materials. When an organism dies, it stops absorbing Carbon-14 from the atmosphere. The existing Carbon-14 atoms in the organism then begin to decay at a predictable rate. This predictable process allows scientists to measure the amount of Carbon-14 remaining and estimate how long it has been since the organism died. This method of using radioactive decay to date ancient artifacts or remains is called radiocarbon dating. Here's how it works:

  • Each radioactive isotope decays at a constant, known rate unique to that isotope.
  • The speed of decay is described by its decay constant.
  • As the isotope decays over time, the proportion of the original material decreases, while the amount of a new form increases.
  • By measuring the remaining isotope, scientists can back-calculate to find out how long the decay process has been happening.
Half-Life
The half-life of a radioactive element is the time required for half of a sample of the element to decay. For Carbon-14, the half-life is approximately 5730 years. This means that every 5730 years, half of the Carbon-14 in a sample will have decayed into a different element. Understanding the concept of half-life is crucial for calculating the age of samples in radiocarbon dating.

The half-life helps scientists to make accurate calculations because the radioactive decay of isotopes is exponential. Some key points about half-life include:

  • Exponential decay means the rate of decay decreases over time, which can be visualized on a graph as a curve.
  • After each half-life period, only half of the remaining radioactive isotopes are left undisinterred.
  • For Carbon-14 dating, knowing the half-life allows scientists to work out how many half-lives have passed since the death of the organism.
  • This allows for the calculation of the time elapsed by measuring what remains of the Carbon-14 in the sample.
Natural Logarithm
The natural logarithm is a mathematical function that is frequently used in calculations involving exponential growth or decay, such as in radiocarbon dating. Denoted as \( \ln(x) \), it is the inverse operation to exponentiation with the base of the constant \( e \), which is approximately equal to 2.71828.

In the context of calculating the age of a bone fragment using Carbon-14 dating, the natural logarithm is used to simplify the equation related to decay. The process involves:

  • Using the equation \( \frac{N(t)}{N_0} = e^{-kt} \) where taking the natural logarithm of both sides helps isolate \( t \).
  • This operation turns the exponential decay equation into a linear one: \( \ln(0.29) = -kt \).
  • Given the logarithmic value, calculations become straightforward, allowing the number of years \( t \) to be easily found.
The natural logarithm simplifies these exponential decay problems into linear equations, making them much easier to solve.
Decay Constant
The decay constant, often represented by the letter \( k \), is crucial in calculations of radioactive decay. It is a measure of the probability per unit time that a given radioactive atom will decay. In relation to Carbon-14, the decay constant is connected to its half-life and helps compute the age of ancient artifacts.

For Carbon-14, the decay constant \( k \) can be computed using the formula \( k = \frac{\ln(2)}{5730} \), due to the known half-life of 5730 years. This value comes out to approximately 0.000121 \( \, \text{years}^{-1} \). Here's how decay constants are used:

  • The decay constant links the radioactive decay rate to the time elapsed, enabling the calculation of the time since an organism's death.
  • Plugging the decay constant into the decay formula helps in deriving the age from the remaining Carbon-14.
  • The detailed decay rate of Carbon-14 is crucial for calculating accurate ages in archaeology and geology.

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Most popular questions from this chapter

CP \({ }^{238} \mathrm{U}\) decays spontaneously by \(\alpha\) emission to \({ }^{234} \mathrm{Th}\). Calculate (a) the total energy released by this process and (b) the recoil velocity of the \({ }^{234}\) Th nucleus. The atomic masses are \(238.050788 \mathrm{u}\) for \({ }^{238} \mathrm{U}\) and \(234.043601 \mathrm{u}\) for \({ }^{234} \mathrm{Th}\).

A thin uniform radioactive coating of an \(a\)-emitter is applied on a metallic sphere of radius ' \(a\) '. The sphere is connected to the ground by means of a thin wire of resistance ' \(R\) ' as shown in figure. Assume that, the sphere is uncharged at \(t=0\), the initial activity of \(a\) - emitter is \(A_{0}\) and the disintegration constant of radioactive material is \(\lambda=3 \sec ^{-1}\). Also assume that all the a-particles emitted go out of the sphere. Find the time (in sec) at which charge on the sphere is maximum. Use \(\ln 2=0.69\) and \(\frac{1}{4 \pi \varepsilon_{0} a}=2 R \lambda\).

Gold, \({ }_{79}^{198} \mathrm{Au}\), undergoes \(\beta^{-}\)decay to an excited state of \({ }_{80}^{198} \mathrm{Hg} .\) If the excited state decays by emission of a \(\gamma\) photon with energy \(0.412 \mathrm{MeV}\), what is the maximum kinetic energy of the electron emitted in the decay? This maximum occurs when the antineutrino has negligible energy. (The recoil energy of the \({ }_{80}^{198} \mathrm{Hg}\) nucleus can be ignored. The masses of the neutral atoms in their ground states are \(197.968225\) u for \({ }_{79}^{198} \mathrm{Au}\) and \(197.966752 \mathrm{u}\) for \(\left.{ }_{80}^{198} \mathrm{Hg} .\right)\)

Radioactive Fallout. One of the problems of in-air testing of nuclear weapons (or, even worse, the use of such weapons!) is the danger of radioactive fallout. One of the most problematic nuclides in such fallout is strontium- \(90\left({ }^{90} \mathrm{Sr}\right)\), which breaks down by \(\beta^{-}\)decay with a half-life of 28 years. It is chemically similar to calcium and therefore can be incorporated into bones and teeth, where, due to its rather long half-life, it remains for years as an internal source of radiation. (a) What is the daughter nucleus of the \({ }^{90} \mathrm{Sr}\) decay? (b) What percentage of the original level of \({ }^{90} \mathrm{Sr}\) is left after 56 years? (c) How long would you have to wait for the original level to be reduced to \(6.25 \%\) of its original value?

A hypothetical decay chain consists of the following elements, \(A, B_{1}, B_{2}\) and \(C\), where \(C\) is stable. The decay constants are \(\lambda, 2 \lambda\) for \(\alpha \& \beta\) decay of \(A\); and \(2 \lambda\), for \(\beta, \alpha\) decays leading to the formation of \(C\). (a) If the atomic and mass number of are \(m, n\) respectively, find the atomic and mass numbers of \(B_{1}, B_{2}\) \& respectively. (b) If at some instant, the number of atoms of \(B_{1}\) reaches a maximum value, find the ratio of the no. of atoms \(A\) to that \(B_{1}\) at this instant. (c) Do the same, when the no. of atoms of \(B_{2}\) reaches maximum, find the ratio of the no. of atoms of \(A\) to that of \(B_{2}\). (d) Find the activity of \(A\) as a function of time.

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