/*! 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 60 The nucleus \({ }_{8}^{15} \math... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The nucleus \({ }_{8}^{15} \mathrm{O}\) has a half-life of \(122.2 \mathrm{~s} ;{ }_{8}^{19} \mathrm{O}\) has a half-life of \(26.9 \mathrm{~s}\). If at some time a sample contains equal amounts of \({ }_{8}^{15} \mathrm{O}\) and \({ }_{8}^{19} \mathrm{O}\), what is the ratio of \({ }_{8}^{15} \mathrm{O}\) to \({ }_{8}^{19} \mathrm{O}\) (a) after 3.0 min and (b) after \(12.0 \mathrm{~min} ?\)

Short Answer

Expert verified
The ratio of isotope \({ }_{8}^{15} \mathrm{O}\) to isotope \({ }_{8}^{19} \mathrm{O}\) is approximately 11:1 after 3 minutes and approximately 4096:1 after 12 minutes.

Step by step solution

01

Convert Times

Given two times 3.0 min and 12.0 min, we need to convert these times into seconds to match the half-life. Therefore, 3.0 mins = 180 sec and 12.0 mins = 720 sec.
02

Calculate Quantity after Decay

Use the decay function \(N = N_0 \times \frac{1}{2}^{\frac{t}{T}}\) for each of \({ }_{8}^{15} \mathrm{O}\) and \({ }_{8}^{19} \mathrm{O}\) to calculate the amount left after 180 sec and then after 720 sec. As the initial amounts of both isotopes are equal, we can set \(N_0 = 1\). This results in a ratio of the isotopes after decay.
03

Ratio after 3.0 min

After 3.0 min (180 sec), the ratio becomes: \(\frac{{}_{8}^{15} \mathrm{O}}{{}_{8}^{19} \mathrm{O}} = \frac{\frac{1}{2}^{\frac{180}{122.2}}}{\frac{1}{2}^{\frac{180}{26.9}}}\) which simplifies to approximately 11:1
04

Ratio after 12.0 min

After 12.0 min (720 sec), repeat the same calculation to find ratio: \(\frac{{}_{8}^{15} \mathrm{O}}{{}_{8}^{19} \mathrm{O}} = \frac{\frac{1}{2}^{\frac{720}{122.2}}}{\frac{1}{2}^{\frac{720}{26.9}}}\) which simplifies to approximately 4096:1

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 91Ó°ÊÓ!

Key Concepts

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

Decay Function
In nuclear physics, the decay function plays a crucial role in understanding how radioactive isotopes lose mass over time. The main formula used is: \[ N = N_0 \times \left( \frac{1}{2} \right)^{\frac{t}{T}} \]Here, \(N\) is the number of remaining nuclei at time \(t\), \(N_0\) is the initial quantity of the substance, \(t\) is the time elapsed, and \(T\) is the half-life of the nuclide. The half-life is the period it takes for half of a radioactive material to decay.
  • This formula hinges on the exponential nature of radioactive decay, indicating that the quantity decreases by half in consistent time intervals.
  • The fractional power \(\left( \frac{t}{T} \right)\) tells us how many half-lives have passed.
  • Knowing the decay function helps in predicting how much of a substance will remain after a certain period.
By applying this formula, students and scientists alike can calculate the amount left of any radioactive isotope after a given time, providing vital information for applications such as carbon dating, nuclear energy production, and medical diagnostics.
Nuclear Physics
Nuclear physics is the field of physics that studies atomic nuclei and their interactions. It sheds light on the fundamental forces governing the nucleus, which includes protons and neutrons.
  • It helps us understand nuclear decay processes like alpha, beta, and gamma decay.
  • It's crucial for applications in nuclear energy, where nuclear reactions produce energy used in power plants.
  • The theories in nuclear physics explain why certain materials are stable and others are not.
This branch of physics also paves the way for technological advances in various fields, including medicine, where it aids in the development of radiation therapies and imaging technologies. By understanding the interactions and transformations within the nucleus, we can manipulate nuclear reactions to obtain desired outcomes, furthering innovation and understanding in the science world.
Radioactive Isotopes
Radioactive isotopes, or radioisotopes, are atoms that contain unstable nuclei, which release energy by emitting radiation during decay. These isotopes have applications across many scientific and practical fields.
  • They are used in medical imaging and treatment, for instance, using isotopes to target and kill cancer cells.
  • In archaeology, radioisotopes help determine the age of ancient objects through methods such as carbon dating.
  • They play a vital role in various research and industrial applications where tracing the movement of materials is necessary.
Each radioisotope decays at its characteristic rate, known as its half-life, making certain isotopes suitable for specific tasks based on how quickly or slowly they decay. For example, isotopes with a short half-life are ideal for medical applications where rapid decay is beneficial, while those with a long half-life might be used in geological research. Understanding the properties and functions of radioactive isotopes allows for their effective and safe utilization in many aspects of modern life.

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

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. \((\mathrm{c})\) 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 \({ }^{4}\) 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.)

We Are Stardust. In 1952 spectral lines of the element technetium-99 \(\left({ }^{99} \mathrm{Tc}\right)\) 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 \({ }^{99} \mathrm{Tc}\) is 200,000 years. (a) For how many half-lives has the \({ }^{99} \mathrm{Tc}\) been in the red giant star if its age is 10 billion years? (b) What fraction of the original \({ }^{99} \mathrm{Tc}\) 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 \({ }^{99} \mathrm{Tc}\) had been 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."

Industrial Radioactivity. Radioisotopes are used in a variety of manufacturing and testing techniques. Wear measurements can be made using the following method. An automobile engine is produced using piston rings with a total mass of \(100 \mathrm{~g},\) which includes \(9.4 \mu \mathrm{Ci}\) of after which the oil is drained and its activity is measured. If the activity of the engine oil is 84 decays/s, how much mass was worn from the piston rings per hour of operation?

Radioactive Tracers. Radioactive isotopes are often introduced into the body through the bloodstream. Their spread through the body can then be monitored by detecting the appearance of radiation in different organs. One such tracer is \({ }^{131} \mathrm{I}\), a \(\beta^{-}\) emitter with a half-life of \(8.0 \mathrm{~d}\). Suppose a scientist introduces a sample with an activity of \(325 \mathrm{~Bq}\) and watches it spread to the organs. (a) Assuming that all of the sample went to the thyroid gland, what will be the decay rate in that gland 24 d (about \(3 \frac{1}{2}\) weeks) later? (b) If the decay rate in the thyroid 24 d later is measured to be 17.0 Bq, what percentage of the tracer went to that gland? (c) What isotope remains after the I-131 decays?

In a diagnostic \(x\) -ray procedure, \(5.00 \times 10^{10}\) photons are absorbed by tissue with a mass of \(0.600 \mathrm{~kg}\). The x-ray wavelength is \(0.0200 \mathrm{nm}\). (a) What is the total energy absorbed by the tissue? (b) What is the equivalent dose in rem?

See all solutions

Recommended explanations on Physics Textbooks

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.