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43.22. Radioactive isotopes used in cancer therapy have a "shelf- life," like pharmaccuticals used in chemotherapy. Just after it has been manufactured in a nuclear reactor, the activity of a sample of "Co is 5000 Ci. When its activity falls below 3500 \(\mathrm{Ci}\) , it is considered too weak a source to use in treatment. You work in the radiology department of a large hospital. One of these "Co sources in your inventory was manufactured on October \(6,2004\) . It is now April \(6,2007\) . Is the source still usable? The half-life of \(^{6}\) Co is 5.271 years.

Short Answer

Expert verified
Yes, the source is still usable as it remains above 3500 Ci.

Step by step solution

01

Understand the Problem

We need to determine if the activity of a radioactive isotope, \(^{60} \text{Co}\), which was 5000 Ci at the time of manufacturing on October 6, 2004, remains above 3500 Ci on April 6, 2007. The half-life of \(^{60} \text{Co}\) is given as 5.271 years.
02

Calculate Time Elapsed

Find the time elapsed between October 6, 2004, and April 6, 2007. This is a period of 2 years and 6 months, which can be expressed as 2.5 years.
03

Understand Radioactive Decay

The decay of radioactive substances follows an exponential decay model. The amount remaining after time \(t\) can be calculated using the decay formula:\[A_t = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}\]where \(A_t\) is the remaining activity, \(A_0\) is the initial activity, \(t\) is the elapsed time, and \(T_{1/2}\) is the half-life.
04

Apply the Decay Formula

Substitute the given values into the decay formula. \[A_t = 5000 \left( \frac{1}{2} \right)^{\frac{2.5}{5.271}}\]Now calculate \(\left( \frac{1}{2} \right)^{\frac{2.5}{5.271}}\) to determine the decay factor.
05

Calculate the Decay Factor

Find the decay factor:\[\left( \frac{1}{2} \right)^{\frac{2.5}{5.271}} \approx 0.7595\]This means that the activity will be reduced to approximately 75.95% of its initial value after 2.5 years.
06

Calculate Remaining Activity

Multiply the initial activity by the decay factor to find the remaining activity:\[A_t = 5000 \times 0.7595 \approx 3797.5 \, \text{Ci}\]
07

Check Usability

Compare the remaining activity with the threshold for usability. Since 3797.5 Ci is greater than 3500 Ci, the source is still usable.

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Key Concepts

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

Half-Life
The concept of half-life is fundamental in understanding radioactive decay. It refers to the time it takes for half of a sample of a radioactive isotope to decay, reducing its activity level by 50%. This process occurs at a consistent rate and is unique to each radioactive isotope. For example, the half-life of Cobalt-60 is 5.271 years.
In practical terms, if you start with a 100 grams sample of Cobalt-60, after 5.271 years, only 50 grams of the isotope will remain radioactive; the rest will have decayed into other elements. The half-life value is critical in fields such as medicine, archaeology, and nuclear physics, allowing for precise calculations in applications like radiotherapy and radiocarbon dating.
  • Predicts when a source will become too weak for use.
  • Helps schedule timely replacement of radioactive material.
Cobalt-60
Cobalt-60 ( ^{60}Co ) is a synthetic radioactive isotope widely used in medical treatments, particularly radiotherapy for cancer. It emits two gamma rays with high energy, which makes it effective in targeting and destroying cancer cells.
Cobalt-60 is produced in nuclear reactors by neutron activation of Cobalt-59. The process involves bombarding stable cobalt atoms with neutrons, making them radioactive. Consequently, ^{60}Co is not found naturally but is manufactured for medical and industrial uses. Its half-life of 5.271 years makes it suitable for extended treatments, but also means careful handling and long-term planning are critical.
  • Known for effectiveness in cancer treatment.
  • Utilized in sterilizing medical equipment.
Exponential Decay
Exponential decay describes how the amount of a substance decreases over time in relation to its half-life. In this context, the activity of a radioactive isotope diminishes by a consistent fraction each half-life period. This behavior is described by the equation: \[ A_t = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]where A_t is the remaining activity at time t, A_0 is the original activity, t is the time elapsed, and T_{1/2} is the half-life.
Using this equation, one can predict the usability of a radioactive source over time and ensure it remains effective for its intended purpose. For example, if a Cobalt-60 source has an initial activity of 5000 Ci, the formula helps determine when it will drop below the usability threshold of 3500 Ci. This is crucial for efficient resource management and safe use in treatments and industrial applications.
  • Allows for precise calculations.
  • Ensures safe and effective use of radioactive materials.
Radiotherapy
Radiotherapy is a medical treatment that uses high-energy radiation to destroy cancer cells. Cobalt-60 is commonly utilized in this process due to its strong gamma radiation, which can penetrate deep into tissues to target tumors.
The precision and reliability of radiotherapy depend heavily on understanding the decay properties of the radioactive sources used, like Cobalt-60. Accurate knowledge of half-life and exponential decay ensures that treatments are delivered at the optimal radiation dose, maximizing effectiveness while minimizing damage to surrounding healthy tissue.
In practice, radiotherapy involves careful planning and dosimetry to align with the decay rates of ^{60}Co . Treatment schedules are meticulously calculated to align with the half-life, ensuring that each session administers the intended level of radiation.
  • Targets and destroys cancer cells.
  • Reduces risks to healthy cells with precise dosages.

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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?

43.55. The atomic mass of \(\frac{25}{12} \mathrm{Mg}\) is 24.985837 \(\mathrm{u}\) , and the atomic mass of 13 \(\mathrm{Al}\) is 24.990429 \(\mathrm{u}\) . (a) Which of these nuclei will decay into the other? (b) What type of decay will occur? Explain how you determined this. (c) How much energy (in MeV) is released in the decay?

43.28. The ratio of \(^{14} \mathrm{C}\) to \(^{12} \mathrm{C}\) in living matter is measured to be \(^{14} \mathrm{C} /^{2} \mathrm{C}=1.3 \times 10^{-12}\) at the present time. A \(12.0-\mathrm{g}\) sample of carbon produces 180 decays/min due to the small amount of \(^{14} \mathrm{C}\) in it. From this information, calculate the half-life of "C.

43.12. The most common isotope of copper is \(\frac{63}{29} \mathrm{Cu}\) . The measured mass of the neutral atom is 62.929601 u. (a) From the measured mass, determine the mass defect, and use it to find the total binding energy and the binding energy per nucleon. (b) Calculare the binding energy from Eq. ( 43.11\()\) . (Why is the fifth term zero? Compare to the result you obtained in part (a). What is the percent difference? What do you conclude about the accuracy of Eq. \((43.11) ?\)

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"

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