/*! 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 12 Radioactive isotopes used in can... [FREE SOLUTION] | 91Ó°ÊÓ

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Radioactive isotopes used in cancer therapy have a "shelf-life," like pharmaceuticals used in chemotherapy. Just after it has been manufactured in a nuclear reactor, the activity of a sample of \({ }^{60} \mathrm{Co}\) is \(5000 \mathrm{Ci}\). 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 \({ }^{60} \mathrm{Co}\) 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 \({ }^{60} \mathrm{Co}\) is \(5.271\) years.

Short Answer

Expert verified
Yes, the source is still usable as its activity is 3765 Ci, which is above the 3500 Ci threshold.

Step by step solution

01

Calculate Elapsed Time

Determine the number of years that have passed from the manufacturing date (October 6, 2004) to the current date (April 6, 2007). Count the full years and fraction thereof. Elapsed time: October 6, 2004 to October 6, 2006 is 2 years. From October 6, 2006 to April 6, 2007 is an additional 6 months, or 0.5 years. Total elapsed time is 2.5 years.
02

Use the Half-Life Formula

The half-life formula is used to calculate the remaining activity of an isotope. The formula is:\[ A = A_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \]where:- \( A \) = remaining activity- \( A_0 \) = initial activity (5000 Ci)- \( t \) = elapsed time (2.5 years)- \( t_{1/2} \) = half-life of the isotope (5.271 years).Substitute the known values into the formula.
03

Calculate Remaining Activity

Substitute the values from Step 2 into the formula to calculate the remaining activity:\[ A = 5000 \times \left(\frac{1}{2}\right)^{\frac{2.5}{5.271}} \]Perform the exponentiation:\[ \left(\frac{1}{2}\right)^{\frac{2.5}{5.271}} \approx 0.753 \]Calculate the activity:\[ A \approx 5000 \times 0.753 \approx 3765 \text{ Ci} \]
04

Compare and Conclude

The minimum activity needed for the isotope to be used in treatment is 3500 Ci. Compare the calculated activity with this threshold: Since the remaining activity (3765 Ci) is greater than 3500 Ci, the source is still usable for treatment.

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Key Concepts

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

Radioactive Decay
Radioactive decay is a natural process where the nucleus of an unstable atom loses energy by emitting radiation. This process changes the atom into a different element or a different isotope of the same element. It involves the transformation of the nucleus as particles such as alpha particles, beta particles, or gamma rays are released. The rate of decay is primarily dependent on the nature of the radioactive material itself, which is measured in terms of activity.
  • Activity: It is the measure of the number of disintegrations or decays per second.
A higher activity indicates a more radioactive sample. While this process might seem random and unpredictable at a microscopic scale, it follows certain patterns statistically over a large number of atoms. This gradual loss of energy through decay helps harness the isotopes for practical applications such as medical treatments, including cancer therapy.
Half-Life Calculation
Half-life is a crucial concept in understanding radioactive decay. It is the time required for half of the radioactive isotopes in a sample to decay. In simpler terms, if you start with a certain amount of radioactive material, after one half-life, only half of it will remain. The calculation of half-life is integral to determining how long a radioactive sample will be useful. By using the half-life formula, one can calculate the remaining activity of a sample over time. The half-life formula is:\[ A = A_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \]
  • \(A\) is the remaining activity.
  • \(A_0\) is the initial activity.
  • \(t\) is the elapsed time since the beginning of the measurement.
  • \(t_{1/2}\) is the half-life of the isotope.
This formula allows for calculating how active a sample is after a given amount of time, considering its half-life.
Cobalt-60
Cobalt-60, often represented as \(^{60}\text{Co}\), is a radioactive isotope of cobalt. It is commonly used in radiation therapy to treat cancer due to its ability to destroy cancerous cells. Cobalt-60 emits gamma rays, which are highly energetic forms of electromagnetic radiation. These gamma rays can penetrate tissues to target cancer cells, making Cobalt-60 a valuable tool in medicine.Here are some key facts about Cobalt-60:
  • Cobalt-60 has a half-life of about 5.271 years. This means it retains its effectiveness for a relatively long period before becoming too weak.
  • It is produced by bombarding cobalt with neutrons in a nuclear reactor.
  • The decay of Cobalt-60 leads to Nickel-60, a stable isotope, providing a predictable transformation path.
Understanding these characteristics of Cobalt-60 helps ensure its safe and effective use in medical treatments.

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Most popular questions from this chapter

Calculate the energy released in the fission reaction \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \rightarrow{ }_{54}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+2{ }_{0}^{1} \mathrm{n} .\) You can ignore the initial kinetic energy of the absorbed neutron. The atomic masses are \({ }_{92}^{235} \mathrm{U}, 235.043923 \mathrm{u} ;{ }_{54}^{140} \mathrm{Xe}, 139.921636 \mathrm{u} ;\) and \({ }_{38}^{94} \mathrm{Sr}, 93.915360 \mathrm{u}\)

The radioactive nuclide \({ }^{199} \mathrm{Pt}\) has a half-life of \(30.8 \mathrm{~min}\) utes. A sample is prepared that has an initial activity of \(7.56 \times 10^{11} \mathrm{~Bq} .\) (a) How many \({ }^{199} \mathrm{Pt}\) nuclei are initially present in the sample? (b) How many are present after \(30.8\) minutes? What is the activity at this time? (c) Repeat part (b) for a time \(92.4\) minutes after the sample is first prepared.

The unstable isotope \(^{40} \mathrm{~K}\) is used for dating rock samples. Its half-life is \(1.28 \times 10^{9} \mathrm{y} .\) (a) How many decays occur per second in a sample containing \(1.66 \times 10^{-6} \mathrm{~g}\) of \({ }^{40} \mathrm{~K} ?\) (b) What is the activity of the sample in curies?

Half lives of two isotopes \(X\) and \(Y\) of a material are known to be \(2 \times 10^{9}\) years and \(4 \times 10^{9}\) years respectively. If a planet was formed with equal number of these isotopes, estimate the current age of the planet, given that currently the material has \(20 \%\) of \(X\) and \(80 \%\) of \(Y\) by number.

A person exposed to fast neutrons receives a radia tion dose of 200 rem on part of his hand, affecting \(25 \mathrm{~g}\) of tissue. The \(\mathrm{RBE}\) of these neutrons is \(10 .\) (a) How many rad did he receive? (b) How many joules of energy did this person receive? (c) Suppose the person received the same rad dosage, but from beta rays with an RBE of \(1.0\) instead of neutrons. How many rem would he have received?

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