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Chapter 19: Problem 45

A rock contains 0.688 \(\mathrm{mg}^{206} \mathrm{Pb}\) for every 1.000 \(\mathrm{mg}^{238} \mathrm{U}\) present. Assuming that no lead was originally present, that all the \(^{206}\mathrm{P}\) formed over the years has remained in the rock, and that the number of nuclides in intermediate stages of decay between \(^{238} \mathrm{U}\) and \(^{206 \mathrm{P}} \mathrm{b}\) is negligible, calculate the age of the rock. (For $38 \mathrm{U}, t_{1 / 2}=4.5 \times 10^{9}$ years.)

Short Answer

Expert verified
The age of the rock is approximately \(3.78 \times 10^9\) years.

Step by step solution

01

Understand the decay formula and variables

The decay formula is given by: \[ N_t = N_0 e^{-\lambda t} \] Where: - \(N_t\) is the number of nuclides remaining at time \(t\). - \(N_0\) is the initial number of nuclides. - \(\lambda\) is the decay constant. - \(t\) is time which we want to find in this problem. After understanding the decay formula and its variables, let's proceed to find the decay constant.
02

Calculate the decay constant (\(\lambda\))

The decay constant is related to the half-life as follows: \[ \lambda = \frac {\ln 2} {t_{1/2}} \] The half-life of \(^{238}\mathrm{U}\) is given as \(t_{1/2}=4.5 \times 10^9 \ \mathrm{years}\). Let's find the decay constant: \[ \lambda = \frac {\ln 2} {4.5 \times 10^9 \ \mathrm{years}} \approx 1.54 \times 10^{-10} \ \mathrm{year}^{-1}\] Now we have the decay constant of \(^{238}\mathrm{U}\). Let's move to the next step.
03

Derive the equation relating the initial and remaining nuclides of \(^{238}\mathrm{U}\) and \(^{206}\mathrm{Pb}\)

We know that: \[ N_t = N_0 e^{-\lambda t} \] Let \(N_U\) and \(N_{Pb}\) be the remaining amounts of \(^{238}\mathrm{U}\) and \(^{206}\mathrm{Pb}\) respectively. Initially, there was no \(^{206}\mathrm{Pb}\) and the amount of \(^{238}\mathrm{U}\) is \(N_0\). So, we can write: \[ N_{Pb} = N_0 - N_U \] As given, the ratio of \(^{206}\mathrm{Pb}\) to \(^{238}\mathrm{U}\) in the rock is \(\frac{0.688}{1}\). Thus, \[ \frac{N_{Pb}}{N_U} = 0.688 \] We can substitute \(N_{Pb}\) from the first equation: \[ \frac {N_0 - N_U}{N_U} = 0.688 \] Now that we have an equation for the initial and remaining \(^{238}\mathrm{U}\) and \(^{206}\mathrm{Pb}\), let's solve for \(N_U\).
04

Calculate the remaining amount of \(^{238}\mathrm{U}\) (\(N_U\))

Let's solve the equation we derived in step 3 for \(N_U\): \[ N_0 - N_U = 0.688 N_U \] \[ N_0 = 1.688 N_U \] \[ N_U = \frac {N_0}{1.688} \] Let's substitute this value in the decay formula for \(N_t\): \[ N_t = N_0 e^{-\lambda t} \] \[ \frac {N_0}{1.688} = N_0 e^{-\lambda t} \] Now, we can solve for time \(t\).
05

Calculate the age of the rock (time \(t\))

We have the decay formula: \[ \frac {N_0}{1.688} = N_0 e^{-\lambda t} \] Divide both sides by \(N_0\): \[ \frac {1}{1.688} = e^{-\lambda t} \] Take the natural logarithm of both sides: \[ \ln \frac {1}{1.688} = -\lambda t \] Now, solve for \(t\): \[ t = \frac {\ln \frac {1}{1.688}}{-\lambda} \] Substitute the value of \(\lambda\): \[ t = \frac {\ln \frac {1}{1.688}}{-1.54 \times 10^{-10} \ \mathrm{year}^{-1}} \approx 3.78 \times 10^9 \ \mathrm{years} \] So, the age of the rock is approximately \(3.78 \times 10^9\) years.

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Most popular questions from this chapter

A positron and an electron can annihilate each other on colliding, producing energy as photons: $$ -_{-1}^{0} e+_{+1}^{0} e \longrightarrow 2_{0}^{0} \gamma $$ Assuming that both g rays have the same energy, calculate the wavelength of the electromagnetic radiation produced.

A reported synthesis of the transuranium element bohrium (Bh) involved the bombardment of berkelium-249 with neon-22 to produce bohrium-267. Write a nuclear reaction for this synthesis. The half-life of bohrium-267 is 15.0 seconds. If 199 atoms of bohrium-267 could be synthesized, how much time would elapse before only 11 atoms of bohrium-267 remain? What is the expected electron configuration of elemental bohrium?

Which of the following statement(s) is (are) true? a. A radioactive nuclide that decays from \(2.00 \times 10^{21}\) atoms to $5.0 \times 10^{20}$ atoms in 16 minutes has a half-life of 8.0 minutes. b. Nuclides with large \(Z\) values are observed to be \(\alpha\) -particle producers. c. As \(Z\) increases, nuclides need a greater proton-to-neutron ratio for stability. d. Those light nuclides that have twice as many neutrons as protons are expected to be stable.

Many transuranium elements, such as plutonium-232 , have very short half- lives. (For \(^{232} \mathrm{Pu}\) , the half-life is 36 minutes.) However, some, like protactinium- 231 (half-life \(=3.34 \times 10^{4}\) years), have relatively long half-lives. Use the masses given in the following table to calculate the change in energy when 1 mole of \(^{232} \mathrm{Pu}\) nuclei and 1 mole of \(^{231} \mathrm{Pa}\) nuclei are each formed from their respective number of protons and neutrons. (Since the masses of \(^{232} \mathrm{Pu}\) and \(^{231} \mathrm{Pa}\) are atomic masses, they each include the mass of the electrons present. The mass of the nucleus will be the atomic mass minus the mass of the electrons.)

Naturally occurring uranium is composed mostly of \(^{238} \mathrm{U}\) and $235 \mathrm{U},\( with relative abundances of 99.28\)\%\( and \)0.72 \%,$ respectively. The half-life for \(^{238} \mathrm{U}\) is \(4.5 \times 10^{9}\) years, and the half-life for 235 \(\mathrm{U}\) is \(7.1 \times 10^{8}\) years. Assuming that the earth was formed 4.5 billion years ago, calculate the relative abundances of the 28 \(\mathrm{U}\) and \(^{235} \mathrm{U}\) isotopes when the earth was formed.
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