/*! 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 72 43.72. An Oceanographic Tracer. ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 43: Problem 72

43.72. An Oceanographic Tracer. Nuclear weapons tests in the 1950 s and 1960 s released significant amounts of radioactive tritium \((3 \mathrm{H}, \text { half-life } 12.3 \text { years) into the atmosphere. The tritium }\) atoms were quickly bound into water molecules and rained out of the air, most of them ending up in the ocean. For any of this tritium-tagged water that sinks below the surface, the amount of time during which it has been isolated from the surface can be calculated by measuring the ratio of the decay product, \(\frac{3}{2} \mathrm{He},\) to the remaining tritium in the water. For example, if the ratio of \(\frac{3}{1}\) He to \(\frac{3}{1} \mathrm{H}\) in a sample of water is \(1 : 1,\) the water has been below the surface for one half-life, or approximately 12 years. This method has provided oceanographers with a convenient way to trace the movements of subsurface currents in parts of the ocean. Suppose that in a particular sample of water, the ratio of \(\frac{3}{2} \mathrm{He}\) to \(\frac{3}{1} \mathrm{H}\) is 4.3 to 1.0 . How many years ago did this water sink below the surface?

Short Answer

Expert verified
The water sank 29.52 years ago.

Step by step solution

01

Understanding the Problem

We are given the ratio of decay product \(\frac{3}{2} \text{He}\) to \(\frac{3}{1} \text{H}\) in a sample of water as 4.3 to 1.0. We need to find out how many half-lives have passed since the water sank below the surface, and then convert that into years.
02

Applying the Decay Ratio

The ratio \(4.3 : 1.0\) means that for every 1 part of remaining tritium \(\frac{3}{1} \text{H}\), there are 4.3 parts of the decay product \(\frac{3}{2} \text{He}\). Since each half-life produces a 1-to-1 ratio (i.e., \(1 : 1\)), we need to calculate how many such half-lives result in a 4.3-to-1 ratio.
03

Calculating Number of Half-lives

Assume that initially there was one part of \(\frac{3}{1} \text{H}\). After one half-life, the ratio will be \(1:1\), after two half-lives it will be \(3:1\) (because \((1+2):1=3:1\)), and after three half-lives, it will be \(7:1\) (because \((1+4.3):1=7:1\)) exponentially growing hence need another way to find.
04

Using the Decimal Model

The ratio \((1 + v):1\) grows as \(1 + 2^n - 1\) gives us \(2^n\) for \(n\) half-lives. For \(4.3 = 2^n - 1\), we must solve for \(n\). So \(2^n = 5.3\). We then log_2 to calculate \(n\rightarrow\frac{5.3/0.301\) = 2.4.
05

Convert Half-lives to Years

Each half-life corresponds to 12.3 years. Now, multiply the number of half-lives 2.4 by 12.3 years to get the total time the water has been isolated: \(2.4 \times 12.3 = 29.52\) years.

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.

Tritium Decay
Tritium, a radioactive isotope of hydrogen, is a key component of studies involving radioactive tracers. Understanding tritium decay is essential for interpreting measurements in fields like oceanography. When tritium (\(^3\text{H}\)) decays, it transforms into helium-3 (\(^3\text{He}\)) through a process known as beta decay.
- In beta decay, a neutron in the tritium nucleus changes to a proton, emitting an electron (beta particle) and an electron antineutrino.- The newly formed \(^3\text{He}\) nucleus consists of two protons and one neutron, making it stable and non-radioactive.
This decay process forms the basis of radioactive tracing techniques, where the ratio of \(^3\text{He}\) to remaining \(^3\text{H}\) acts as a clock to trace the age of water or other materials. Interpreting this ratio helps us calculate how long the radioactive material has been undergoing decay.
Oceanographic Measurements
In oceanography, radioactive tracers like tritium are invaluable for measuring oceanic processes. Tritium's involvement in oceanographic measurements lies in its ability to time the movement and origin of water masses. By studying water samples tagged with tritium, scientists can trace how long these water bodies have been isolated from surface exchange.
- **Subsurface Currents Tracking**: After tritium-laden water enters the ocean, it can sink and become part of a subsurface current. By analyzing the tritium to \(^3\text{He}\) ratio, oceanographers can reveal how long the water has been traveling since it became isolated from the atmosphere.- **Water Mass Identification**: Identifying water masses through radioactive tracers aids in understanding ocean dynamics and climate patterns. This involves tracking how water with characteristic tritium levels circulates globally.
Additionally, the distinct tritium signature from nuclear tests provides a unique marker for ocean water, enhancing our understanding of subsurface ocean currents and their interaction with climate systems.
Half-life Calculations
The concept of half-life is fundamental in calculating the time elapsed since a radioactive substance began to decay. A half-life is the time required for half of a quantity of radioactive substance to decay into its daughter elements. For tritium, the half-life is 12.3 years.
- **Understanding Half-life**: When a material undergoes radioactive decay, each half-life period reduces the quantity of the original radionuclide by half. After one half-life, its amount reduces to 50%, after two half-lives to 25%, and so on.
- **Example Calculation**: In the given example, if the ratio of \(^3\text{He}\) to \(^3\text{H}\) is 4.3:1, this indicates that over several half-lives, more \(^3\text{He}\) has accumulated compared to \(^3\text{H}\). Using logarithmic methods or a decay formula, you can calculate that approximately 2.4 half-lives have passed.
- **Total Time Elapsed**: By multiplying the number of half-lives that have passed by the tritium half-life (12.3 years), you derive the total time. Here, this results in approximately 29.52 years since the water was last at the surface, providing insight into the historical movement of this water body.

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

43.36. To Sean or Not to Scan? It has become popular for some people to have yearly whole-body scans (CT scans, formerly called CAT scans using \(x\) rays, just to see if they detect anything suspicious. A number of medical people have recently questioned the advisability of such scans, due in part to the radiation they impart. Typically, one such scan gives a dose of 12 \(\mathrm{mSv}\) , applied to the whole body. By contrast, a chest \(\mathbf{x}\) ray typically administers 0.20 \(\mathrm{mSv}\) to only 5.0 \(\mathrm{kg}\) of tissue. How many chest \(\mathrm{x}\) rays would deliver the same total amount of energy to the body of \(75-\mathrm{kg}\) person as one whole-body scan?

43.8. Calculate (a) the total binding energy and (b) the binding energy per nucleon of \(^{12} \mathrm{C}\) (c) What percent of the rest mass of this nucleus is its total binding energy?

43.65. We Are Stardust. In 1952 spectral lines of the element technetium- 99 ( \(^{99} \mathrm{Tc} )\) were discovered in a red giant star. Red giants are very old stars, often around 10 billion years old, and near the end of their lives. Technetium has no stable isotopes, and the half-life of \(^{99} \mathrm{Te}\) is \(200,000\) years. (a) For how many half-lives has the \(^{99}\) Te been in the red-giant star if its age is 10 billion years? (b) What fraction of the original "Te would be left at the end of that time? This discovery was extremely important because it provided convincing evidence for the theory (now essentially known to be true) that most of the atoms heavier than hydrogen and helium were made inside of stars by thermonuclear fusion and other nuclear processes. If the \(^{99} \mathrm{Tchadbeen}\) part of the star since it was born, the amount remaining after 10 billion years would have been so minute that it would not have been detectable. This knowledge is what led the late astronomer Carl Sagan to proclaim that "we are stardust"

43.52. Comparison of Energy Released per Gram of Fuel. (a) When gasoline is burned, it releases \(1.3 \times 10^{8} \mathrm{J}\) of energy per gallon \((3.788 \mathrm{L})\) . Given that the density of gasoline is 737 \(\mathrm{kg} / \mathrm{m}^{3}\) , express the quantity of energy released in \(\mathrm{J} / \mathrm{g}\) of fuel. (b) During fission, when a neutron is absorbed by a \(^{235} \mathrm{U}\) nucleus, about 200 \(\mathrm{MeV}\) of energy is released for each nucleus that undergoes fission. Express this quantity in \(\mathrm{J} / \mathrm{g}\) of fuel. (c) In the proton-proton chain that takes place in stars like our sun, the overall fusion reaction can be summarized as six protons fusing to form one 4 \(\mathrm{He}\) nucleus with two leftover protons and the liberation of 26.7 \(\mathrm{MeV}\) of energy. The fuel is the six protons. Express the energy produced here in units of \(\mathrm{J} / \mathrm{g}\) of fuel. Notice the huge difference between the two forms of nuclear energy, on the one hand, and the chemical energy from gasoline, on the other (d) Our sun produces energy at a measured rate of \(3.86 \times 10^{26} \mathrm{W}\) . If its mass of \(1.99 \times 10^{30} \mathrm{kg}\) were all gasoline, how long could it last before consuming all its fuel? (Historical note: Before the discovery of nuclear fusion and the vast amounts of energy it releases, scientists were confused. They knew that the earth was at least many millions of years old, but could not explain how the sun could survive that long if its energy came from chemical burning.)

43.35. A nuclear chemist receives an accidental radiation dose of 5.0 Gy from slow neutrons \((\mathrm{RBE}=4.0) .\) What does she receive in rad, rem, and \(\mathrm{J} / \mathrm{kg} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Rotational Dynamics

Read Explanation

Thermodynamics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Translational Dynamics

Read Explanation

Engineering 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.