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The radioactive potassium- 40 isotope decays to argon- 40 with a half-life of \(1.2 \times 10^{9}\) yr. (a) Write a balanced equation for the reaction. (b) A sample of moon rock is found to contain 18 percent potassium40 and 82 percent argon by mass. Calculate the age of the rock in Years.

Short Answer

Expert verified
The age of the moon rock is approximately \(4.18 \times 10^{9}\) years.

Step by step solution

01

Write the balanced radioactive decay reaction

A balanced radioactive decay reaction conserves mass numbers and atomic numbers. Potassium-40 to Argon-40 involves beta decay, where an electron (or beta particle) is emitted from the nucleus. The balanced equation is: \(^{40}_{19}K \rightarrow ^{40}_{18}Ar + ^{0}_{-1}\beta + \gamma\). Here \(\gamma\) represents the gamma radiation released during decay.
02

Understand the concept of Half-life

Half-life is the time required for half of the atoms in a radioactive sample to decay. In this exercise, the half-life of potassium-40 is given to be \(1.2 \times 10^{9}\) years.
03

Use the decay relation

We know that the remaining fraction of the isotope after time t is given by \(f = \frac{1}{2}^{(t/T)}\), where T is the half-life of the isotope. From the question, we know that 82% of the original potassium-40 has decayed, meaning that 18% remains. Hence, \(f = 0.18\).
04

Solve for the time (t) of decay

Substitute \(f = 0.18\) and \(T = 1.2 \times 10^{9}\) years into the decay relation, then solve for \(t\). On solving this equation, \(t = -1.2 \times 10^{9} \ln(0.18) / \ln(0.5)\) years. The value of \(t\) is approximately \(4.18 \times 10^{9}\) years.

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Key Concepts

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

Beta Decay
Beta decay is a fundamental process that helps us understand radioactive transformations. It occurs when a radioactive isotope, like Potassium-40, transforms into another element. During beta decay, a neutron in the nucleus converts into a proton.
This process results in the emission of a beta particle, which is essentially an electron.
  • The atomic number of the element changes because the neutron becomes a proton.
  • For Potassium-40, the atomic number changes from 19 (Potassium) to 18 (Argon).
  • The mass number remains unchanged, as both the proton and neutron have similar masses.
In the balanced equation for potassium decay: \[^{40}_{19}\text{K} \rightarrow ^{40}_{18}\text{Ar} + ^{0}_{-1}\beta + \gamma \]The gamma radiation \(\gamma\), a type of electromagnetic radiation, is also emitted during this transformation but does not affect atomic or mass numbers.
Half-life
Half-life is a critical concept in understanding radioactive decay. It represents the time it takes for half of the radioactive isotopes in a sample to decay.
The idea is straightforward but crucial for calculations in radioactive processes.
  • For Potassium-40, the half-life is about \(1.2 \times 10^9\) years.
  • When determining how much of a radioactive substance remains, half-life serves as a key factor.
  • As time progresses in units of the half-life, the remaining quantity of the isotope drops by half each period.
In calculations, the half-life helps determine periods of decay and proportion of remaining substance, allowing scientists and geologists to calculate time intervals effectively.
Radioactive Isotopes
Radioactive isotopes are atoms that are unstable and naturally transform over time. They undergo decay, a process which stabilizes the atom by emitting particles.
Potassium-40 is one of these isotopes.
  • Unlike stable isotopes, radioactive isotopes break down into different elements.
  • For example, Potassium-40 decays into Argon-40 through a series of nuclear changes.
  • This transformation includes the emission of particles like beta particles and gamma rays.
Radioactive isotopes like Potassium-40 are valuable in various scientific fields, including archaeology and geology, for dating purposes, because they act as natural clocks.
Age Determination
Age determination using radioactive isotopes is a reliable method to date ancient objects or geological formations. Potassium-40 is commonly used in this method.
The process involves measuring the proportions of the parent isotope (in this case, Potassium-40) to the daughter product (here, Argon-40).
  • By analyzing the ratio of these elements, scientists can determine how much time has passed since the rock or object was formed.
  • The half-life of the isotope plays a crucial role in these calculations, by predicting how much of the parent isotope should remain after a given time period.
  • In the exercise, we calculated that about \(4.18 \times 10^9\) years have passed since the formation of a moon rock based on the decay of Potassium-40 to Argon-40.
Tools like these are essential for constructing historical and geological timelines, enhancing our understanding of Earth's history and the timelines of various events.

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Most popular questions from this chapter

The quantity of a radioactive material is often measured by its activity (measured in curies or millicuries) rather than by its mass. In a brain scan procedure, a 70 -kg patient is injected with \(20.0 \mathrm{mCi}\) of \({ }^{99 \mathrm{~m}} \mathrm{Tc}\) which decays by emitting \(\gamma\) -ray photons with a halflife of \(6.0 \mathrm{~h}\). Given that the \(\mathrm{RBE}\) of these photons is 0.98 and only two-thirds of the photons are absorbed by the body, calculate the rem dose received by the patient. Assume all of the \({ }^{99 \mathrm{~m}}\) Tc nuclei decay while in the body. The energy of a gamma photon is \(2.29 \times 10^{-14} \mathrm{~J}\).

Which of the following poses a greater health hazard: a radioactive isotope with a short half-life or a radioactive isotope with a long half-life? Explain. [Assume same type of radiation \((\alpha\) or \(\beta)\) and comparable energetics per particle emitted.

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The half-life of \({ }^{27} \mathrm{Mg}\) is \(9.50 \mathrm{~min}\). (a) Initially there were \(4.20 \times 10^{12}{ }^{27} \mathrm{Mg}\) nuclei present. How many \({ }^{27}\) Mg nuclei are left 30.0 min later? (b) Calculate the \({ }^{27} \mathrm{Mg}\) activities \((\) in \(\mathrm{Ci})\) at \(t=0\) and \(t=30.0 \mathrm{~min}\) (c) What is the probability that any one \({ }^{27} \mathrm{Mg}\) nucleus decays during a 1 -s interval? What assumption is made in this calculation?

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