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The cloth shroud from around a mummy is found to have a \({ }^{14} \mathrm{C}\) activity of \(9.7\) disintegrations per minute per gram of carbon as compared with living organisms that undergo \(16.3\) disintegrations per minute per gram of carbon. From the half-life for \({ }^{14} \mathrm{C}\) decay, \(5715 \mathrm{yr}, \mathrm{cal}-\) culate the age of the shroud.

Short Answer

Expert verified
The age of the cloth shroud is approximately 3597 years.

Step by step solution

01

Determine the ratio of remaining \({ }^{14} \mathrm{C}\) activities

We start by calculating the ratio of the given activity for the cloth shroud to the activity for living organisms: \[ \frac{N}{N_0} = \frac{9.7}{16.3} \]
02

Calculate the ratio of remaining \({ }^{14} \mathrm{C}\) amounts

Now that we have the ratio of the activities, we can simplify it to find the ratio of the remaining \({ }^{14} \mathrm{C}\) amounts: \[ \frac{N}{N_0} = 0.59508 \]
03

Use the half-life formula

Since we have the value of \(\frac{N}{N_0}\), we can use the half-life formula to find the age of the shroud: \[0.59508 = (1/2)^{t/5715} \]
04

Take the logarithm of both sides

To solve for \(t\), we should take the logarithm of both sides. Using a base of 2 will simplify the equation: \[ \log_2(0.59508) = \frac{t}{5715} \cdot \log_2(1/2) \]
05

Solve for \(t\)

Now, we can solve for \(t\): \[t = (\log_2(0.59508) \times 5715 ) / (\log_2(1/2)) \] \[t \approx 3596.7 \] The age of the cloth shroud is approximately 3597 years.

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Key Concepts

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

Carbon-14 Decay
Carbon-14 decay is a fascinating natural process that helps us understand the age of ancient organic materials through radiocarbon dating. Carbon-14, or \(^{14}C\), is a radioactive isotope of carbon that is incorporated into living organisms. While alive, creatures maintain a constant amount of \(^{14}C\) in their bodies, equal to the levels in the atmosphere.
However, when an organism dies, it no longer absorbs \(^{14}C\). Instead, the isotope begins to decay, turning into nitrogen-14 over time. This decay is measured by detecting the number of disintegrations per minute per gram of carbon that occur.
By comparing the present \(^{14}C\) activity of a sample to that of living creatures, scientists can estimate how long it has been since the organism's death, using this reduction in \(^{14}C\) as a natural clock.
Half-life Calculation
The concept of half-life is crucial for understanding radioactive decay and, by extension, radiocarbon dating. A half-life is the time required for half of the radioactive substance to decay. In the case of Carbon-14, the half-life is approximately 5715 years.
When calculating the age of a sample, like a mummy's shroud, using half-life involves comparing the ratio of the current \(^{14}C\) activity to the original activity of living organisms.
The formula for the age \(t\) of a sample can be represented by: \[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^{t/T} \]where \(N/N_0\) is the ratio of the remaining activity, \(t\) is the age, and \(T\) is the half-life. By solving for \(t\), we can determine the approximate age of the sample.
Radioactive Decay
Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. There are several types of radioactive decay, but for Carbon-14, the process involves beta decay, where a neutron in the nucleus of \(^{14}C\) transforms into a proton, creating a nitrogen-14 atom.
This transformation is random, but the statistical probability of decay remains constant, thus making the decay rate predictable over a large number of atoms.
Radiocarbon dating takes advantage of this predictability by using the decay rate of \(^{14}C\) to estimate the age of organic compounds. The exponential nature of radioactive decay means that although some \(^{14}C\) atoms will decay quickly, others will take longer—hence the need for statistical methods and half-life calculations to determine the age of ancient materials accurately.

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Most popular questions from this chapter

Cobalt-60 is a strong gamma emitter that has a half-life of \(5.26 \mathrm{yr}\). The cobalt-60 in a radiotherapy unit must be replaced when its radioactivity falls to \(75 \%\) of the original sample. (a) If an original sample was purchased in June 2006, when will it be necessary to replace the cobalt-60? (b) How can you store cobalt-60 so that it is safe to handle?

An experiment was designed to determine whether an aquatic plant absorbed iodide ion from water. Iodine\(131\left(t_{1 / 2}=8.02\right.\) days) was added as a tracer, in the form of iodide ion, to a tank containing the plants. The initial activity of a \(1.00-\mu \mathrm{L}\) sample of the water was 214 counts per minute. After 30 days the level of activity in a \(1.00-\mu \mathrm{L}\) sample was \(15.7\) counts per minute. Did the plants absorb iodide from the water? Explain.

Calculate the binding energy per nucleon for the following nuclei: (a) \({ }^{14} \mathrm{~N}\) (nuclear mass, \(13.999234\) amu); (b) \({ }^{48} \mathrm{Ti}\) (nuclear mass, \(47.935878\) amu); (c) xenon-129 (atomic mass, \(128.904779\) amu).

How much energy must be supplied to break a single \({ }^{21}\) Ne nucleus into separated protons and neutrons if the nucleus has a mass of \(20.98846\) amu? What is the nuclear binding energy for 1 mol of \({ }^{21}\) Ne?

A laboratory rat is exposed to an alpha-radiation source whose activity is \(14.3 \mathrm{mCi}\). (a) What is the activity of the radiation in disintegrations per second? In becquerels? (b) The rat has a mass of \(385 \mathrm{~g}\) and is exposed to the radiation for \(14.0 \mathrm{~s}\), absorbing \(35 \%\) of the emitted alpha particles, each having an energy of \(9.12 \times 10^{-13} \mathrm{~J} .\) Calculate the absorbed dose in millirads and grays. (c) If the RBE of the radiation is \(9.5\), calculate the effective absorbed dose in mrem and Sv.

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