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\(\textbf{Radiation Treatment of Prostate Cancer}\). In many cases, prostate cancer is treated by implanting 60 to 100 small seeds of radioactive material into the tumor. The energy released from the decays kills the tumor. One isotope that is used (there are others) is palladium (\(^1$$^0$$^3\)Pd), with a half-life of 17 days. If a typical grain contains 0.250 g of \(^1$$^0$$^3\)Pd, (a) what is its initial activity rate in Bq, and (b) what is the rate 68 days later?

Short Answer

Expert verified
(a) Initial activity: \(6.89 \times 10^{14}\) Bq. (b) Activity after 68 days: \(4.30 \times 10^{13}\) Bq.

Step by step solution

01

Understanding Half-life

The half-life of an isotope is the time taken for half of the radioactive atoms to decay. In this problem, the half-life of palladium (\(^{103}\text{Pd}\)) is given as 17 days.
02

Calculating Initial Activity

Activity \( A \) of a radioactive substance is related to its decay constant \( \lambda \) and the number of active nuclei \( N \) by the formula: \( A = \lambda N \). The decay constant is related to the half-life \( T_{1/2} \) by \( \lambda = \frac{\ln 2}{T_{1/2}} \). Calculate \( \lambda \): \( \lambda = \frac{\ln 2}{17~\text{days}} = 0.0408~\text{day}^{-1} \).
03

Converting Mass to Moles

Find the number of moles in 0.250 g of \(^{103}\text{Pd}\). Using the molar mass of palladium as 103 g/mol, the number of moles \( n \) is: \( n = \frac{0.250~\text{g}}{103~\text{g/mol}} = 0.00243~\text{mol} \).
04

Finding the Number of Atoms

Calculate the number of atoms \( N \) using Avogadro's number \( 6.022 \times 10^{23}~\text{atoms/mol} \): \( N = 0.00243~\text{mol} \times 6.022 \times 10^{23}~\text{atoms/mol} = 1.46 \times 10^{21}~\text{atoms} \).
05

Calculating Initial Activity in Bq

Use the formula \( A = \lambda N \) to calculate the initial activity: \[ A = 0.0408~\text{day}^{-1} \times 1.46 \times 10^{21}~\text{atoms} = 5.95 \times 10^{19}~\text{decays/day} \]. Convert this to Bq (1 Bq = 1 decay/s): \( A = \frac{5.95 \times 10^{19}}{86400} \approx 6.89 \times 10^{14}~\text{Bq} \).
06

Calculating Activity after 68 Days

Use the decay formula: \( A_t = A_0 \cdot e^{-\lambda t} \). Here \( t = 68~\text{days} \), \( \lambda = 0.0408~\text{day}^{-1} \) and \( A_0 = 6.89 \times 10^{14}~\text{Bq} \). \[ A_{68} = 6.89 \times 10^{14} \cdot e^{-0.0408 \times 68} \]. Calculate \( A_{68} \): use \( e^{-0.0408 \times 68} \approx 0.0625 \), so \( A_{68} \approx 6.89 \times 10^{14} \times 0.0625 \approx 4.30 \times 10^{13}~\text{Bq} \).

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Key Concepts

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

Half-life
The half-life of a radioactive isotope is crucial for understanding how long the isotope will remain active and how fast it will decay. It is defined as the time required for half of the radioactive atoms in a sample to decay. For palladium-103, the isotope used in prostate cancer treatment, this half-life is 17 days. This means that every 17 days, the quantity of radioactive palladium-103 will reduce to half its initial amount.

Half-life is not dependent on the amount of substance nor on external conditions such as temperature or pressure. Instead, it is a unique property of each radioactive isotope.

The relationship between half-life and decay constant is given by:
  • \( ext{Decay constant} \, (\lambda) = \frac{\ln 2}{T_{1/2}}\)
  • For palladium-103, \(\lambda = \frac{\ln 2}{17} \approx 0.0408 \text{ day}^{-1}\).
Activity Rate
The activity rate of a radioactive substance measures how many decays occur per second or more commonly per day. It's typically given in Becquerels (Bq), where 1 Bq represents one decay per second.

The initial activity of a radioactive isotope is calculated using its decay constant \(\lambda\) and the number of radioactive atoms \(N\):
  • \(A = \lambda N\)
For example, when using palladium-103 in prostate cancer, we calculate the initial activity by determining the number of atoms in a given mass and multiplying by \(\lambda\).

Over time, the activity decreases exponentially, described by the equation:
  • \(A_t = A_0 \cdot e^{-\lambda t}\)
telling us how the activity lessens as the radioactive material decays.
Radioactive Isotopes
Radioactive isotopes are atoms with unstable nuclei that release energy in the form of radiation as they decay into more stable forms. These isotopes are used widely in medical applications, including cancer treatment.

Palladium-103 is one such isotope, commonly used in brachytherapy for prostate cancer. This treatment involves implanting tiny seeds containing the isotope directly into or near the tumor.

The choice of isotope depends on its half-life and radiation emission type. A suitable radioactive isotope for brachytherapy would have:
  • A half-life long enough to remain effective through treatment.
  • Radiation that can effectively penetrate tumor cells without harming nearby healthy tissue.
Prostate Cancer Treatment
Prostate cancer treatment often utilizes a technique known as brachytherapy, which involves placing radioactive 'seeds' within the prostate to treat the tumor effectively. These seeds emit radiation that kills cancer cells while minimizing damage to surrounding healthy tissue.

This procedure is minimally invasive and is chosen for its precision in targeting tumors directly. Radioactive isotopes like palladium-103 are preferred due to their effective radiation and half-life, ensuring the cancer cells receive a sufficient dose over the treatment period. Benefits include:
  • Short recovery times.
  • Effectiveness for early-stage cancers.
  • Lower risk of side effects compared to traditional therapies.
Understanding the decay process and how isotopes like palladium-103 release energy is essential for optimizing treatment strategies and improving patient outcomes.

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Most popular questions from this chapter

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 \(^6$$^0\)Co is 5000 Ci. When its activity falls below 3500 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 \(^6$$^0\)Co sources in your inventory was manufactured on October 6, 2011. It is now April 6, 2014. Is the source still usable? The half-life of \(^6$$^0\)Co is 5.271 years.

\(\textbf{Comparison of Energy Released per Gram of Fuel.}\) (a) When gasoline is burned, it releases 1.3 \(\times\) 10\(^8\) J of energy per gallon (3.788 L). Given that the density of gasoline is 737 kg/m\(^3\), express the quantity of energy released in J/g of fuel. (b) During fission, when a neutron is absorbed by a \(^2$$^3$$^5\)U nucleus, about 200 MeV of energy is released for each nucleus that undergoes fission. Express this quantity in J/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 \(^4\)He nucleus with two leftover protons and the liberation of 26.7 MeV of energy. The fuel is the six protons. Express the energy produced here in units of J/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\(^2$$^6\) W. If its mass of 1.99 \(\times\) 10\(^3$$^0\) 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.)

Measurements indicate that 27.83% of all rubidium atoms currently on the earth are the radioactive \(^8$$^7\)Rb isotope. The rest are the stable \(^8$$^5\)Rb isotope. The half-life of \(^8$$^7\)Rb is 4.75 \(\times\) 10\(^1$$^0\) y. Assuming that no rubidium atoms have been formed since, what percentage of rubidium atoms were \(^8$$^7\)Rb when our solar system was formed 4.6 \(\times\) 10\(^9\) y ago?

At the beginning of Section 43.7 the equation of a fission process is given in which \(^2$$^3$$^5\)U is struck by a neutron and undergoes fission to produce \(^1$$^4$$^4\)Ba, \(^8$$^9\)Kr, and three neutrons. The measured masses of these isotopes are 235.043930 u (\(^2$$^3$$^5\)U), 143.922953 u (\(^1$$^4$$^4\)Ba), 88.917631 u (\(^8$$^9\)Kr), and 1.0086649 u (neutron). (a) Calculate the energy (in MeV) released by each fission reaction. (b) Calculate the energy released per gram of \(^2$$^3$$^5\)U, in MeV/g.

Hydrogen atoms are placed in an external magnetic field. The protons can make transitions between states in which the nuclear spin component is parallel and antiparallel to the field by absorbing or emitting a photon. What magnetic- field magnitude is required for this transition to be induced by photons with frequency 22.7 MHz?

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