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Chapter 21: Problem 39

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鈥攈ence 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

The naturally occurring radioactive decay series that begins with \({ }_{92}^{235} \mathrm{U}\) stops with formation of the stable \({ }_{82}^{20} \mathrm{~Pb}\) nucleus. The decays proceed through a series of alpha-particle and beta-particle emissions. How many of each type of emission are involved in this series?

It has been suggested that strontium-90 (generated by nuclear testing) deposited in the hot desert will undergo radioactive decay more rapidly because it will be exposed to much higher average temperatures. (a) Is this a reasonable suggestion? (b) Does the process of radioactive decay have an activation energy, like the Arrhenius behavior of many chemical reactions \({ }^{\infty} 0\) (Section \(\left.14.5\right) ?\) Discuss.

What particle is produced during the following decay processes: (a) sodium-24 decays to magnesium-24; (b) mercury-188 decays to gold-188; (c) iodine-122 decays to xenon-122; (d) plutonium-242 decays to uranium-238?

The energy from solar radiation falling on Earth is \(1.07 \times 10^{16} \mathrm{~kJ} / \mathrm{min}\). (a) How much loss of mass from the Sun occurs in one day from just the energy falling on Earth? (b) If the energy released in the reaction $$ { }^{235} \mathrm{U}+{ }_{0^{n}}{ }_{\mathbf{n}} \longrightarrow{ }^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_{0}^{1} \mathrm{n} $$ \(\left({ }^{235} \mathrm{U}\right.\) nuclear mass, \(234.9935 \mathrm{amu} ;{ }^{141}\) Ba nuclear mass, \(140.8833 \mathrm{amu} ;{ }^{92} \mathrm{Kr}\) nuclear mass, \(91.9021 \mathrm{amu}\) ) is taken as typical of that occurring in a nuclear reactor, what mass of uranium- 235 is required to equal \(0.10 \%\) of the solar energy that falls on Earth in \(1.0\) day?

Write balanced nuclear equations for the following transformations: (a) gold-191 undergoes electron capture; (b) gold-201 decays to a mercury isotope; (c) gold198 undergoes beta decay; (d) gold-188 decays by positron emission.
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