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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

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 g, which includes \(9.4 \mu Ci\) of \(^{59}Fe\) whose half-life is 45 days. The engine is test-run for 1000 hours, 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?

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