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The half-life for the process \({ }^{238} \mathrm{U} \longrightarrow{ }^{206} \mathrm{~Pb}\) is \(4.5 \times 10^{9} \mathrm{yr}\). A mineral sample contains \(75.0 \mathrm{mg}\) of \({ }^{239} \mathrm{U}\) and \(18.0 \mathrm{mg}\) of \({ }^{206} \mathrm{~Pb}\). What is the age of the mineral? Energy Changes in Nuclear Reactions (Section 21.6)

Short Answer

Expert verified
The age of the mineral is approximately \(1.14\times10^9\) years.

Step by step solution

01

Understand the radioactive decay formula

We will use the following radioactive decay formula to determine the time t that has elapsed since the decay process began: \[ N(t) = N_0 e^{-λt} \] where, - \(N(t)\) is the amount of radioactive substance remaining at time t - \(N_0\) is the initial amount of radioactive substance - \(λ\) is the decay constant - \(t\) is the time that has passed
02

Calculate the decay constant using half-life

We are given the half-life, \(T_{1/2}\), of the decay process as \(4.5\times10^9\) years. The decay constant, \(λ\), can be calculated using the following formula: \[ λ = \frac{\ln 2}{T_{1/2}} \] Plug in the value of \(T_{1/2}\) to find the decay constant, \(λ\): \[ λ = \frac{\ln 2}{4.5\times10^9} \]
03

Determine the initial amount of uranium-238

We are given that the initial amount of uranium-239 in the mineral is \(75.0\text{ mg}\). Since uranium-238 decays into lead-206, we can assume that at the beginning of the decay process, there was no lead-206 in the sample. So, the initial amount of uranium-238 would be equal to the sum of the current amounts of uranium-238 and lead-206: \[ N_0 = 75.0\text{ mg} + 18.0\text{ mg} = 93.0\text{ mg} \]
04

Calculate the time that has passed

Now we need to determine the time, \(t\), that has passed since the beginning of the decay process. We can rearrange the radioactive decay formula to solve for \(t\): \[ t = \frac{\ln(\frac{N_0}{N(t)})}{λ} \] Plug in the values for \(N_0\), \(N(t)\), and \(λ\), and solve for \(t\): \[ t = \frac{\ln(\frac{93.0}{75.0})}{\frac{\ln 2}{4.5\times10^9}} \]
05

Calculate the age of the mineral

After evaluating the expression, we get the age of the mineral: \[ t \approx 1.14 \times 10^9 \text{ years} \] So, the age of the mineral is approximately \(1.14\times10^9\) years.

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Key Concepts

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

half-life
The half-life is a crucial concept in understanding radioactive decay. It refers to the time required for half of a radioactive substance to decay. Half-life is constant, unaffected by external conditions like temperature or pressure. Therefore, it provides a reliable measure of the decay rate.
To calculate the half-life, we use the formula:
  • \[T_{1/2} = \frac{\ln 2}{λ}\]
where \(T_{1/2}\) is the half-life and \(λ\) is the decay constant.
For the decay process of Uranium-238 into Lead-206, the given half-life is \(4.5 \times 10^9\) years. This means that every \(4.5 \times 10^9\) years, half of the Uranium-238 in a sample will decay into Lead-206. Understanding the half-life allows us to estimate the age of geological samples by examining the ratio of parent (Uranium-238) to daughter (Lead-206) isotopes.
decay constant
The decay constant, denoted as \(λ\), is an essential parameter in the study of radioactive decay. It characterizes the rate at which a substance disintegrates over time. The decay constant is linked to the half-life by the formula:
  • \[λ = \frac{\ln 2}{T_{1/2}}\]
The decay constant is measured in inverse time units, such as \(\text{yr}^{-1}\), indicating how quickly the decay process happens.
For Uranium-238, with a half-life of \(4.5 \times 10^9\) years, the decay constant can be calculated by plugging in the half-life value. After calculating, you will find that \(λ\) is a very small number, highlighting the slow nature of Uranium-238's decay process. This slow decay is what makes Uranium-238 valuable for studying the age of rocks and the Earth itself. The smaller the decay constant, the slower the decay rate, and consequently, the longer it takes for a significant amount of U-238 to decay.
Uranium-238
Uranium-238 is a naturally occurring isotope of uranium and one of the most important elements when it comes to dating geological formations. It undergoes radioactive decay to form Lead-206 in a series of steps, known as the Uranium decay series.
Some interesting points about Uranium-238 include:
  • It has a very long half-life of \(4.5 \times 10^9\) years, which makes it ideal for dating the Earth's oldest rocks.
  • During decay, it emits alpha particles, which is a relatively common type of decay among heavy elements.
  • Uranium-238 is plentiful in the Earth's crust, making it accessible for scientific studies.
In the context of the exercise, Uranium-238's decay to Lead-206 is used to estimate the age of a mineral sample. By analyzing the ratio of Uranium-238 to Lead-206 in the sample, and knowing the half-life, scientists can calculate how much time has passed since the mineral formation. This process is a fundamental technique in geochronology, the study of Earth's age and the timing of geological events.

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Most popular questions from this chapter

Indicate the number of protons and neutrons in the following nuclei: (a) \({ }_{53}^{129} \mathrm{I}\), (b) \({ }^{138} \mathrm{Ba}\), (c) neptunium-237.

Each statement that follows refers to a comparison between two radioisotopes, \(A\) and \(X\). Indicate whether each of the following statements is true or false, and why. (a) If the half-life for \(\mathrm{A}\) is shorter than the half-life for \(\mathrm{X}, \mathrm{A}\) has a larger decay rate constant. (b) If \(X\) is "not radioactive," its half-life is essentially zero. (c) If A has a half-life of \(10 \mathrm{yr}\), and \(\mathrm{X}\) has a half-life of \(10,000 \mathrm{yr}\), A would be a more suitable radioisotope to measure processes occurring on the 40 -yr time scale.

Decay of which nucleus will lead to the following products: (a) bismuth- 211 by beta decay; (b) chromium 50 by positron emission; (c) tantalum-179 by electron capture; (d) radium-226 by alpha decay?

A wooden artifact from a Chinese temple has a \({ }^{14} \mathrm{C}\) activity of \(38.0\) counts per minute as compared with an activity of \(58.2\) counts per minute for a standard of zero age. From the halflife for \({ }^{14} \mathrm{C}\) decay, \(5715 \mathrm{yr}\), determine the age of the artifact.

Nuclear scientists have synthesized approximately \(1600 \mathrm{nu}-\) clei not known in nature. More might be discovered with heavy-ion bombardment using high-energy particle accelerators. Complete and balance the following reactions, which involve heavy-ion bombardments: (a) \({ }_{3}^{6} \mathrm{Li}+{ }_{28}^{56} \mathrm{Ni} \longrightarrow\) ? (b) \({ }_{20}^{40} \mathrm{Ca}+{ }_{96}^{248} \mathrm{Cm} \longrightarrow{ }_{62}^{147} \mathrm{Sm}+\) ? (c) \({ }_{38}^{88} \mathrm{Sr}+{ }_{26}^{81} \mathrm{Kr} \longrightarrow{ }_{26}^{116} \mathrm{Pd}+\) ? (d) \({ }_{20}^{40} \mathrm{Ca}+{ }_{92}^{23 \mathrm{~S}} \mathrm{U} \longrightarrow{ }_{30}^{70} \mathrm{Zn}+4{ }_{0}^{1} \mathrm{n}+2\) ?

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