/*! 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 37 The radionuclide \({ }^{64} \mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 42: Problem 37

The radionuclide \({ }^{64} \mathrm{Cu}\) has a half-life of \(12.7 \mathrm{~h}\). If a sample contains \(5.50 \mathrm{~g}\) of initially pure \({ }^{64} \mathrm{Cu}\) at \(t=0,\) how much of it will decay between \(t=14.0 \mathrm{~h}\) and \(t=16.0 \mathrm{~h} ?\)

Short Answer

Expert verified
0.23 grams of \\({}^{64}Cu\\) decayed.

Step by step solution

01

Determine Initial Decay Constant

The decay constant \( \lambda \) is calculated using the formula \( \lambda = \frac{\ln(2)}{\text{half-life}} \). The half-life provided is 12.7 hours, so\[ \lambda = \frac{\ln(2)}{12.7} \approx 0.0546 \text{ hr}^{-1}. \]
02

Calculate Remaining Material at t = 14.0 h

We use the formula \( N(t) = N_0 e^{-\lambda t} \), where \( N_0 = 5.50 \, \text{g} \). At \( t = 14.0 \, \text{h} \):\[ N(14) = 5.50 \, \text{g} \times e^{-0.0546 \times 14} \approx 3.38 \, \text{g}. \]
03

Calculate Remaining Material at t = 16.0 h

Using the same decay formula, at \( t = 16.0 \, \text{h} \):\[ N(16) = 5.50 \, \text{g} \times e^{-0.0546 \times 16} \approx 3.15 \, \text{g}. \]
04

Determine Amount Decayed Between t = 14.0 h and t = 16.0 h

The amount that decayed is the difference between the material present at \( t = 14.0 \, \text{h} \) and \( t = 16.0 \, \text{h} \):\[ \Delta N = N(14) - N(16) = 3.38 \, \text{g} - 3.15 \, \text{g} = 0.23 \, \text{g}. \]
05

Conclusion

The amount of \( {}^{64} \text{Cu} \) that decayed between \( t = 14.0 \, \text{h} \) and \( t = 16.0 \, \text{h} \) is 0.23 grams.

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.

Understanding Half-Life
The half-life of a radioactive substance is the time it takes for half of the radioactive atoms to decay. It acts as a clock to measure how quickly a substance loses its radioactivity. A key point is that the half-life remains constant, no matter how much substance you start with or how much is left. For example, with a half-life of 12.7 hours, in 12.7 hours half of the ^{64}Cu in the sample will have decayed.
This concept is useful in many fields, from archaeology to medicine.
  • Archaeologists use it to date artifacts through carbon dating.
  • Medical professionals use it to calculate safe dosages of radiopharmaceuticals.
Every radioactive isotope has its own unique half-life. This characteristic makes half-life calculations critical when working with different radionuclides.
The Decay Constant
The decay constant, denoted by \( \lambda, \) is a probability measure of the decay process. It describes how frequently the individual atoms decayed.
Calculated using the formula \( \lambda = \frac{\ln(2)}{\text{half-life}}, \) it links to the half-life by showing the decay rate.
For ^{64}Cu with a half-life of 12.7 hours, this gives a decay constant of approximately 0.0546 \( \text{hr}^{-1}. \)
  • A larger decay constant means a faster decay rate.
  • This measure is critical for predicting how much of a substance remains over time.
Understanding the decay constant helps in comprehending how rapidly a substance loses its radioactivity over a given period.
Exploring Exponential Decay
Exponential decay describes how the quantity of a radioactive substance decreases over time. The substance reduces at a rate proportional to its current value, creating a curve that represents rapid early loss, slowing down over time.
The mathematical representation is \( N(t) = N_0 e^{-\lambda t}, \) where \( N_0 \) is the initial amount.
  • The equation uses the decay constant \( \lambda \) to predict remaining material after a time \( t. \)
  • In the exercise, you see this calculation predicting the amounts left at different times.
  • Exponential decay is foundational in understanding how substances lose mass and radioactivity.
This concept applies to any radioactive substance and is vital for applications like nuclear power generation or dating ancient biological materials.
Radionuclide Basics
A radionuclide, also known as a radioactive isotope, is an atom with an unstable nucleus. This instability causes the nucleus to release energy in the form of radiation until it becomes stable. \(^{64}Cu \) is an example where it emits radiation until it transforms into a stable form.
Key characteristics include:
  • Radionuclides occur naturally or can be artificially produced.
  • Each one has a unique decay mode, resulting in different types of radiation.
  • They are used in medicine, industry, and scientific research.
Understanding radionuclides helps in comprehending phenomena such as background radiation and the workings of nuclear reactors. It aids in solving the exercise by acknowledging how much of a radionuclide like \(^{64}Cu \) decays in a given time.

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 radionuclide \({ }^{11} \mathrm{C}\) decays according to $${ }^{11} \mathrm{C} \rightarrow{ }^{11} \mathrm{~B}+\mathrm{e}^{+}+\nu, \quad T_{1 / 2}=20.3 \mathrm{~min}$$ The maximum energy of the emitted positrons is \(0.960 \mathrm{MeV}\). (a) Show that the disintegration energy \(Q\) for this process is given by $$Q=\left(m_{\mathrm{C}}-m_{\mathrm{B}}-2 m_{\mathrm{e}}\right) c^{2}$$ where \(m_{\mathrm{C}}\) and \(m_{\mathrm{B}}\) are the atomic masses of \({ }^{11} \mathrm{C}\) and \({ }^{11} \mathrm{~B}\), respectively, and \(m_{e}\) is the mass of a positron. (b) Given the mass values \(m_{\mathrm{C}}=11.011434 \mathrm{u}, m_{\mathrm{B}}=11.009305 \mathrm{u},\) and \(m_{\mathrm{e}}=\) \(0.0005486 \mathrm{u},\) calculate \(Q\) and compare it with the maximum energy of the emitted positron given above. (Hint: Let \(\mathbf{m}_{\mathrm{C}}\) and \(\mathbf{m}_{\mathrm{B}}\) be the nuclear masses and then add in enough electrons to use the atomic masses.)

The radioactive nuclide \({ }^{99}\) 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 \({ }^{99} \mathrm{Mo},\) 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 excited-state \({ }^{99} \mathrm{Tc}\) are located in the tumor at that time?

What is the binding energy per nucleon of the americium isotope \({ }_{95}^{244} \mathrm{Am} ?\) Here are some atomic masses and the neutron mass. $$\begin{array}{lr}{ }_{95}^{244} \mathrm{Am} & 244.064279 \mathrm{u} \\\\\mathrm{n} & 1.008665 \mathrm{u}\end{array} { }^{1} \mathrm{H} \quad 1.007825 \mathrm{u}$$

Large radionuclides emit an alpha particle rather than other combinations of nucleons because the alpha particle has such a stable, tightly bound structure. To confirm this statement, calculate the disintegration energies for these hypothetical decay processes and discuss the meaning of your findings: (a) \({ }^{235} \mathrm{U} \rightarrow{ }^{232} \mathrm{Th}+{ }^{3} \mathrm{He},\) (b) \({ }^{235} \mathrm{U} \rightarrow{ }^{231} \mathrm{Th}+{ }^{4} \mathrm{He}\) (c) \({ }^{235} \mathrm{U} \rightarrow{ }^{230} \mathrm{Th}+{ }^{5} \mathrm{He}\) The needed atomic masses are $$ \begin{array}{llll} { }^{232} \mathrm{Th} & 232.0381 \mathrm{u} & { }^{3} \mathrm{He} & 3.0160 \mathrm{u} \\ { }^{231} \mathrm{Th} & 231.0363 \mathrm{u} & { }^{4} \mathrm{He} & 4.0026 \mathrm{u} \\ { }^{230} \mathrm{Th} & 230.0331 \mathrm{u} & { }^{5} \mathrm{He} & 5.0122 \mathrm{u} \\ { }^{235} \mathrm{U} & 235.0429 \mathrm{u} & & \end{array} $$

Cancer cells are more vulnerable to \(\mathrm{x}\) and gamma radiation than are healthy cells. In the past, the standard source for radiation therapy was radioactive \({ }^{60} \mathrm{Co},\) which decays, with a half-life of \(5.27 \mathrm{y},\) into an excited nuclear state of \({ }^{60} \mathrm{Ni}\). That nickel isotope then immediately emits two gamma-ray photons, each with an approximate energy of \(1.2 \mathrm{MeV}\). How many radioactive \({ }^{60} \mathrm{Co}\) nuclei are present in a \(6000 \mathrm{Ci}\) source of the type used in hospitals? (Energetic particles from linear accelerators are now used in radiation therapy.)
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Linear Momentum

Read Explanation

Fields in Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Radiation

Read Explanation

Medical 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.