/*! 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 54 The unstable isotope \({ }^{40} ... [FREE SOLUTION] | 91Ó°ÊÓ

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The unstable isotope \({ }^{40} \mathrm{~K}\) is used to date rock samples. Its half-life is \(1.28 \times 10^{8}\) years. (a) How many decays occur per second in a sample containing \(6.00 \times 10^{-6} \mathrm{~g}\) of \({ }^{40} \mathrm{~K} ?\) (b) What is the activity of the sample curies?

Short Answer

Expert verified
(a) 1.553 decays per second. (b) 4.198 x 10^-11 curies.

Step by step solution

01

Understanding Decay Constant

To solve the problem, first, we determine the decay constant \(\lambda\) using the isotope's half-life. The formula for decay constant is \(\lambda = \frac{\ln(2)}{T_{1/2}}\), where \(T_{1/2} = 1.28 \times 10^8\, \text{years}\).
02

Convert Half-life to Seconds

Since activity is given in decays per second, convert the half-life from years to seconds: \(1 \text{ year} = 3.15 \times 10^7 \text{ seconds}\). Thus, \(T_{1/2} = 1.28 \times 10^8 \times 3.15 \times 10^7 = 4.032 \times 10^{15}\, \text{seconds}\).
03

Calculate Decay Constant

Now calculate \(\lambda\): \(\lambda = \frac{\ln(2)}{4.032 \times 10^{15}} = 1.72 \times 10^{-16}\, \text{sec}^{-1}\).
04

Calculate Number of K-40 Atoms

Determine the number of \(^{40}K\) atoms: Use Avogadro's number \(6.022 \times 10^{23} \, \text{atoms/mol}\) and the molar mass of \(^{40}K\), which is approximately \(40\, \text{g/mol}\). The number of moles \(n = \frac{6.00 \times 10^{-6}\, \text{g}}{40\, \text{g/mol}} = 1.5 \times 10^{-7}\, \text{mol}\). The number of atoms \(N = 1.5 \times 10^{-7}\, \text{mol} \times 6.022 \times 10^{23}\, \text{atoms/mol} = 9.033 \times 10^{16}\, \text{atoms}\).
05

Calculate Decays per Second (Activity)

The activity \(A\) is the rate of decay, given by \(A = \lambda N\). Thus, \(A = 1.72 \times 10^{-16} \times 9.033 \times 10^{16} = 1.553 \, \text{decays per second}\).
06

Convert Activity to Curies

One curie is \(3.7 \times 10^{10}\, \text{decays/second}\). The activity in curies is \(\frac{1.553}{3.7 \times 10^{10}} = 4.198 \times 10^{-11}\, \text{curies}\).

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Key Concepts

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

Isotope Half-Life
Isotope half-life is a fundamental concept in understanding radioactive decay and is used to determine the age of materials, like rocks and fossils. Half-life, denoted as \(T_{1/2}\), is the time required for half of a given amount of radioactive isotopes to decay. This means that if you start with a certain amount of an isotope, after one half-life, only half of the original amount will remain.
For example, the half-life of \(^{40}K\) is given as \(1.28 \times 10^{8}\) years. This means if you start with a sample containing 100% \(^{40}K\), only 50% of it will remain after \(1.28 \times 10^{8}\) years, 25% after another \(1.28 \times 10^{8}\) years, and so on. This exponential decay happens because each atom has a fixed probability of decaying within a given time frame.
Understanding half-life helps in dating objects and materials by calculating how many half-lives have occurred. Scientists apply this in fields such as archaeology, geology, and other sciences to trace historical timelines and make discoveries about our planet and beyond.
Decay Constant Calculation
The decay constant, represented by \(\lambda\), is essential for describing how quickly a radioactive substance decays. It connects directly to the half-life using the relation \(\lambda = \frac{\ln(2)}{T_{1/2}}\).
To practically use this, first, convert the half-life into seconds if given in different units, since decay constants are typically measured per second. In the calculation for the exercise, \(^{40}K\)’s half-life is initially in years. Multiply by the number of seconds in a year, \(3.15 \times 10^{7}\), to convert it. This results in \(T_{1/2} = 4.032 \times 10^{15}\) seconds.
Once in seconds, apply the formula to get the decay constant: \(\lambda = \frac{\ln(2)}{4.032 \times 10^{15}} \approx 1.72 \times 10^{-16} \text{ sec}^{-1}\). The decay constant provides a way to quantify the rate of decay, helping in calculating the activity of a sample.
Radioactivity Units
Radioactivity is often expressed in terms of activity, which is the number of decays occurring each second. Different units are used to measure radioactivity, with the curie and the becquerel being the most common.
The becquerel (Bq) is the SI unit of radioactivity, defined as one decay per second. In contrast, the curie (Ci) is an older unit based on the activity of 1 gram of radium-226, originally named after Marie Curie and Pierre Curie. It equals \(3.7 \times 10^{10}\) decays per second.
  • 1 Bq = 1 decay per second
  • 1 Ci = \(3.7 \times 10^{10}\) decays per second
When performing calculations, converting the activity from decays per second to curies helps understand the scale of radioactivity relative to known quantities. For example, if the activity of \(^{40}K\) is calculated as 1.553 decays per second, converting it to curies involves dividing by \(3.7 \times 10^{10}\), which results in \(4.198 \times 10^{-11}\) curies.

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Most popular questions from this chapter

At the beginning of Section \(30.6,\) a fission process is illustrated in which \({ }^{235} \mathrm{U}\) is struck by a neutron and undergoes fission to produce \({ }^{144} \mathrm{Ba},{ }^{89} \mathrm{Kr},\) and three neutrons. The measured masses of these isotopes are \(235.043930 \mathrm{u}\left({ }^{235} \mathrm{U}\right), 143.922953 \mathrm{u}\left({ }^{144} \mathrm{Ba}\right), 88.917630 \mathrm{u}\) \(\left({ }^{89} \mathrm{Kr}\right),\) and \(1.0086649 \mathrm{u}\) (neutron). (a) Calculate the energy (in \(\mathrm{MeV}\) ) released by each fission reaction. (b) Calculate the energy released per gram of \({ }^{235} \mathrm{U},\) in \(\mathrm{MeV} / \mathrm{g} .\)

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