/*! 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 64 co The isotope \({ }^{40} \mathr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 42: Problem 64

co The isotope \({ }^{40} \mathrm{~K}\) can decay to either \({ }^{40} \mathrm{Ca}\) or \({ }^{40} \mathrm{Ar}\); assume both decays have a half-life of \(1.26 \times 10^{9} \mathrm{y}\). The ratio of the Ca produced to the Ar produced is \(8.54 / 1=8.54\). A sample originally had only \({ }^{40} \mathrm{~K}\). It now has equal amounts of \({ }^{40} \mathrm{~K}\) and \({ }^{40} \mathrm{Ar} ;\) that is, the ratio of \(\mathrm{K}\) to Ar is \(1 / 1=1\). How old is the sample? (Hint: Work this like other radioactive-dating problems, except that this decay has two products.)

Short Answer

Expert verified
The sample is approximately 2.25 billion years old.

Step by step solution

01

Understanding the decay process

The isotope \(^{40} \text{K}\) decays into \(^{40} \text{Ar}\) and \(^{40} \text{Ca}\). We have two separate decay paths with the same half-life of \(1.26 \times 10^{9} \text{ years}\). The given ratio of \(^{40} \text{Ca}\) to \(^{40} \text{Ar}\) is \(8.54:1\). The problem states that the remaining amounts of \(^{40} \text{K}\) and \(^{40} \text{Ar}\) are equal, meaning \([K]/[Ar] = 1\).
02

Setting up the decay equations

Let \(N_0\) be the initial amount of \(^{40} \text{K}\), \(N_K\) the remaining \(^{40} \text{K}\), \(N_{Ar}\) the amount of \(^{40} \text{Ar}\), and \(N_{Ca}\) the amount of \(^{40} \text{Ca}\). We know: \(N_K + N_{Ar} + N_{Ca} = N_0\). Also, we have the decay ratios \(N_{Ar} = N_K\) and \(N_{Ca} / N_{Ar} = 8.54\).
03

Expressing decay in terms of half-life

The radioactive decay formula is \(N(t) = N_0 e^{-\lambda t}\) where \(\lambda = \frac{\ln(2)}{\text{half-life}}\). Simplifying this for both paths:1. \(N_K = N_0/2\) for the remaining potassium, since \(N_K = N_{Ar}\).2. For \(N_{Ca} = 8.54 N_{Ar}\), insert into: \(N_0 = N_K + N_{Ar} + (8.54)N_{Ar}\).
04

Solving for time

Substitute in the total initial amount: \(N_0 = N_{Ar} + N_{Ar} + 8.54 N_{Ar} = 10.54 N_{Ar}\). This means \(N_{Ar}/N_0 = 1/10.54\) and the decay equation \(N(t) = N_0 e^{-\lambda t}\) becomes:\[\frac{N_{Ar}}{N_0} = e^{-\lambda t} = \frac{1}{10.54}\]Solving \(e^{-\lambda t} = \frac{1}{10.54}\), use \(\lambda = \frac{\ln(2)}{1.26 \times 10^9}\).
05

Calculate the time

Using the natural logarithm to solve:\[-\lambda t = \ln(\frac{1}{10.54})\]\[t = \frac{-\ln(\frac{1}{10.54})}{\lambda}\]Substitute \(\lambda = \frac{\ln(2)}{1.26 \times 10^9}\):\[t = \frac{-\ln(\frac{1}{10.54})}{\frac{\ln(2)}{1.26 \times 10^9}}\]Evaluating gives \(t \approx 2.25 \times 10^9 \text{ years}\).
06

Conclusion: Determine the Sample Age

The calculated age of the sample based on the decay equations and using the half-life is \(2.25 \times 10^9 \text{ years}\). This indicates how long ago the sample contained only \(^{40} \text{K}\) before decaying into \(^{40} \text{Ar}\) and \(^{40} \text{Ca}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Isotopic Decay
In science, isotopic decay describes a natural process where an unstable atomic nucleus loses energy by emitting radiation. Over time, isotopes, which are variants of elements with the same number of protons but different numbers of neutrons, can transform into a different element or a different isotope of the same element. In our exercise, the decay process involves the transition of the isotope \(^{40} \text{K}\) (potassium) into either \(^{40} \text{Ar}\) (argon) or \(^{40} \text{Ca}\) (calcium).
  • During decay, we track the amount of original isotope remaining compared to the amount changed into the products over time.
  • For \(^{40} \text{K}\), there are two decay pathways, both leading to different isotopes, with a specific ratio of \(^{40} \text{Ca}\)/\(^{40} \text{Ar}\) = 8.54.
  • This decay is fundamental to dating techniques because it provides a natural clock, indicating how much time has passed since the isotope began to transform.
Understanding isotopic decay is essential, as it allows us to calculate the age of archaeological samples and geological formations. This decay process's known rates help us derive significant historical data.
Potassium-Argon Dating
Potassium-Argon dating is a specific method of radiometric dating that leverages the decay of potassium (\(^{40} \text{K}\)) to argon (\(^{40} \text{Ar}\)) to determine the age of a sample. This method is particularly useful in dating volcanic rocks because argon, as a gas, escapes from molten rock, leaving no argon in freshly formed rock. Any argon we find in the rock must have developed there after the rock was solidified.
  • In our context, this dating method is slightly complex because \(^{40} \text{K}\) also decays into \(^{40} \text{Ca}\).
  • The method requires understanding how the initial amount of \(^{40} \text{K}\) decayed into both \(^{40} \text{Ar}\) and \(^{40} \text{Ca}\) over time.
  • By measuring the ratio of \(^{40} \text{Ar}\) to the remaining \(^{40} \text{K}\), and considering the known decay pathway to \(^{40} \text{Ca}\), we can accurately calculate how long the decay has been occurring.
This approach relies on precise measurement and careful calculations, making it a robust tool for geological and archaeological dating.
Half-Life Calculation
The concept of half-life is crucial in understanding radioactive dating. Half-life is defined as the time it takes for half of the radioactive isotope present in a sample to decay into its products. It provides a consistent rate at which isotopic decay occurs, thus acting as a reliable temporal measure.
  • The half-life of \(^{40} \text{K}\) in our exercise is \(1.26 \times 10^9\) years, meaning after that duration, half of the original \(^{40} \text{K}\) converts to \(^{40} \text{Ar}\) and \(^{40} \text{Ca}\).
  • In our example, we assumed equal amounts of \(^{40} \text{K}\) and \(^{40} \text{Ar}\), informing us about the elapsed time since no \(^{40} \text{Ar}\) was initially present.
  • We use the formula \(N(t) = N_0 e^{-\lambda t}\), where \(\lambda = \frac{\ln(2)}{\text{half-life}}\), to calculate the age of the sample, providing the exact time the decay process started.
Half-life calculates the sample's age correctly, ensuring the reliability of isotopic dating methods. This consistent decay rate is why radioactive dating can be so precise when determining geological and historical timelines.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The half-life of a particular radioactive isotope is \(6.5 \mathrm{~h}\). If there are initially \(48 \times 10^{19}\) atoms of this isotope, how many remain at the end of \(26 \mathrm{~h}\) ?

The radioactive nuclide \({ }^{99} \mathrm{Tc}\) can be injected into a patient's bloodstream in order to monitor the blood flow, measure the blood volume, or find a tumor, among other goals. The nuclide is produced in a hospital by a "cow" containing \({ }^{99}\) Mo, a radioactive nuclide that decays to \({ }^{99} \mathrm{Tc}\) with a half-life of \(67 \mathrm{~h} .\) Once a day, the cow is "milked" for its \({ }^{99} \mathrm{Tc}\), which is produced in an excited state by the \({ }^{99} \mathrm{Mo} ;\) the \({ }^{99} \mathrm{Tc}\) de-excites to its lowest energy state by emitting a gamma-ray photon, which is recorded by detectors placed around the patient. The de-excitation has a half- life of \(6.0 \mathrm{~h}\). (a) By what process does \({ }^{99}\) Mo decay to \({ }^{99} \mathrm{Tc}\) ? (b) If a patient is injected with an \(8.2 \times 10^{7}\) Bq sample of \({ }^{99} \mathrm{Tc}\), how many gamma-ray photons are initially produced within the patient each second? (c) If the emission rate of gamma-ray photons from a small tumor that has collected \({ }^{99} \mathrm{Tc}\) is 38 per second at a certain time, how many excitedstate \({ }^{99} \mathrm{Tc}\) are located in the tumor at that time?

A radium source contains \(1.00 \mathrm{mg}\) of \({ }^{226} \mathrm{Ra}\), which decays with a half-life of \(1600 \mathrm{y}\) to produce \({ }^{222} \mathrm{Rn}\), a noble gas. This radon isotope in turn decays by alpha emission with a half-life of \(3.82 \mathrm{~d}\). If this process continues for a time much longer than the half-life of \({ }^{222} \mathrm{Rn}\), the \({ }^{222}\) Rn decay rate reaches a limiting value that matches the rate at which \({ }^{222} \mathrm{Rn}\) is being produced, which is approximately constant because of the relatively long half-life of \({ }^{226} \mathrm{Ra}\). For the source under this limiting condition, what are (a) the activity of \({ }^{226} \mathrm{Ra},(\mathrm{b})\) the activity of \({ }^{222} \mathrm{Rn}\), and \((\mathrm{c})\) the total mass of \({ }^{222} \mathrm{Rn}\) ?

A \(1.00 \mathrm{~g}\) sample of samarium emits alpha particles at a rate of 120 particles/s. The responsible isotope is \({ }^{147} \mathrm{Sm}\), whose natural abundance in bulk samarium is \(15.0 \%\). Calculate the half-life.

What is the mass excess \(\Delta_{1}\) of \({ }^{1} \mathrm{H}\) (actual mass is \(\left.1.007825 \mathrm{u}\right)\) in (a) atomic mass units and (b) \(\mathrm{MeV} / \mathrm{c}^{2}\) ? What is the mass excess \(\Delta_{\mathrm{n}}\) of a neutron (actual mass is \(1.008665 \mathrm{u}\) ) in (c) atomic mass units and (d) \(\mathrm{MeV} / \mathrm{c}^{2} ?\) What is the mass excess \(\Delta_{120}\) of \({ }^{120} \mathrm{Sn}\) (actual mass is \(119.902197 \mathrm{u})\) in (e) atomic mass units and (f) \(\mathrm{MeV} / \mathrm{c}^{2}\) ?
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Fields in Physics

Read Explanation

Medical Physics

Read Explanation

Dynamics

Read Explanation

Solid State Physics

Read Explanation

Nuclear Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.