Problem 50
Question
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)
Step-by-Step Solution
Verified Answer
The age of the mineral is approximately \(1.14\times10^9\) years.
1Step 1: 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
2Step 2: 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, \(λ\):
\[ λ = \frac{\ln 2}{4.5\times10^9} \]
3Step 3: 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:
\[ N_0 = 75.0\text{ mg} + 18.0\text{ mg} = 93.0\text{ mg} \]
4Step 4: 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 \(t\):
\[ t = \frac{\ln(\frac{93.0}{75.0})}{\frac{\ln 2}{4.5\times10^9}} \]
5Step 5: 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.
Key Concepts
half-lifedecay constantUranium-238
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:
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.
To calculate the half-life, we use the formula:
- \[T_{1/2} = \frac{\ln 2}{λ}\]
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:
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.
- \[λ = \frac{\ln 2}{T_{1/2}}\]
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:
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.
Other exercises in this chapter
Problem 48
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
View solution Problem 49
Potassium-40 decays to argon-40 with a half-life of \(1.27 \times 10^{9} \mathrm{yr}\), What is the age of a rock in which the mass ratio of \({ }^{40} \mathrm{
View solution Problem 57
How much energy must be supplied to break a single \({ }^{21} \mathrm{Ne}\) nucleus into separated protons and neutrons if the nucleus has a mass of \(20.98846\
View solution Problem 58
The atomic masses of hydrogen-2 (deuterium), helium-4, and lithium-6 are \(2.014102 \mathrm{amu}_{2} 4.002602 \mathrm{amu}\), and \(6.0151228\) amu, respectivel
View solution