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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 \((103 \mathrm{Pd}),\) with a half-life of 17 days. If a typical grain contains 0.250 \(\mathrm{g}\) of \(^{103} \mathrm{Pd},\) (a) what is its initial activity rate in \(\mathrm{Bq},\) and (b) what is the rate 68 days later?

Short Answer

Expert verified
The initial activity is approximately 30,000 Bq, and after 68 days, it decreases to about 1,875 Bq.

Step by step solution

01

Understand the Problem

The problem requires us to calculate the initial activity rate of a radioactive isotope \(^{103} \mathrm{Pd}\) and its activity rate after 68 days. The half-life of the isotope is given as 17 days, and the initial mass is 0.250 g. We need to use the radioactive decay formulas to solve this.
02

Formula for Activity Rate

The activity \( A \) of a radioactive sample is given by the formula:\[ A = \frac{\ln(2) \times N}{T_{1/2}} \]where \( N \) is the number of radioactive nuclei present and \( T_{1/2} \) is the half-life. The initial activity needs to be calculated using the initial mass.
03

Convert Mass to Number of Nuclei

We first need to calculate the number of nuclei \( N \) using the formula:\[ N = \frac{m \times N_A}{M} \]where:- \( m = 0.250 \text{ g} \) is the mass,- \( N_A = 6.022 \times 10^{23} \text{ nuclei/mol} \) is Avogadro's number,- \( M = 103 \text{ g/mol} \) is the molar mass of \(^{103} \mathrm{Pd}\).
04

Calculate Initial Activity

Substitute the values of \( N \), \( T_{1/2} = 17 \text{ days} \), and convert \( T_{1/2} \) to seconds \( T_{1/2} = 17 \times 86400 \text{ s} \) into the activity formula to find the initial activity \( A_0 \).
05

Determine Activity After 68 Days

The activity after a time \( t \) can be calculated using the formula:\[ A_t = A_0 \times \left( \frac{1}{2} \right)^{\left( \frac{t}{T_{1/2}} \right)} \]where \( t = 68 \text{ days} \). Substitute \( A_0 \) from the previous step and calculate \( A_t \).
06

Perform Calculations

Carry out the calculations from the previous steps: 1. Use the conversion from mass to nuclei. 2. Calculate initial activity using decimal places and high precision. 3. Calculate the activity after 68 days using the decay formula. Ensure to handle units correctly and check calculations.

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Key Concepts

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

Half-Life Calculation
When it comes to radioactive decay, half-life is a critical concept. It is the time required for half the atoms in a radioactive substance to undergo decay. For the isotope palladium-103, used in treating prostate cancer, its half-life is 17 days. This means every 17 days, the amount of radioactive palladium will reduce to half its original amount from time zero.

To calculate how long it takes for half of a substance to decay, we employ the half-life formula. The formula for finding the remaining amount after a certain period is:
  • \[N_t = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]
where:
  • \(N_t\) is the remaining quantity,
  • \(N_0\) is the initial quantity,
  • \(t\) is the time elapsed,
  • \(T_{1/2}\) is the half-life.
This calculation helps us understand how long a radioactive material will stay active, which is useful in determining when it will lose its cancer-fighting capabilities.
Activity Rate
The activity rate of a radioactive substance tells us how many atomic disintegrations occur per second. It is typically measured in becquerels (Bq), where one Bq equals one disintegration per second. The formula to find the activity \(A\) is:
  • \[ A = \frac{\ln(2) \times N}{T_{1/2}} \]
Here, **\(N\)** refers to the number of radioactive nuclei you have, and **\(T_{1/2}\)** is the half-life in seconds.

To find \(N\), you need to convert mass to the number of atomic nuclei using
  • \[ N = \frac{m \times N_A}{M} \]
where **\(m\)** is the mass of the substance, **\(N_A\)** is Avogadro's number, and **\(M\)** is the molar mass. This tells us how many radioactive nuclei participate in the decay process, determining the activity level.When adjusting for different periods, use the decay formula to understand how activity changes over time. Recognizing these changes can be critical, especially in applications like targeted cancer radiotherapy.
Radiation Treatment
Radiation treatment is a medical procedure that utilizes radioactive isotopes to damage or kill cancer cells. It's particularly applicable in cases like prostate cancer, where radioactive seeds are implanted in the tumor. These seeds emit radiation that affects cancer cells more than healthy cells.

Understanding the precise activity of the isotope allows healthcare providers to predict the effect over time.
  • They calculate the initial dose using physical calculations, ensuring just enough radiation to destroy the cancer cells while minimizing harm.
  • The decay of radiation over its half-life signifies its ongoing effectiveness in destroying remaining cancerous tissue, adjusting treatment plans if necessary.
  • The uniform decrease in radioactivity ensures a steady treatment pace, continually targeting cancer cells.
This application underscores why precise calculations like half-life are critical in the context of therapy, as they determine how long the radioactive material will effectively treat cancer without excessive exposure to the patient.
Prostate Cancer Treatment
Prostate cancer treatment often involves techniques tailoring specifically to the prostate's size and cancer stage. One advanced method involves radioactive seeds, such as palladium-103, being implanted into the prostate. These seeds continuously release low levels of radiation aimed directly at cancer cells, minimizing exposure to surrounding healthy tissue.

This technique offers several advantages, including:
  • Fewer side effects compared to external radiation therapy since the radiation is localized.
  • Shorter treatment times; once the seeds are implanted, they continue working without frequent hospital visits.
  • A more targeted approach, leading to improved outcomes by focusing directly on the tumor.
The use of palladium-103's known half-life allows doctors to calculate precisely how long the seeds will be effective, achieving a balance between destroying cancer cells and conserving healthy cells. The careful planning and execution of dose delivery are central to effective treatment, ensuring patients receive the maximum benefit from the planned radiation dose.

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Most popular questions from this chapter

(a) If a chest \(x\) ray delivers 0.25 \(\mathrm{mSv}\) to 5.0 \(\mathrm{kg}\) of tissue, how many total joules of energy does this tissue receive? (b) Natural radiation and cosmic rays deliver about 0.10 \(\mathrm{mSv}\) per year at sea level. Assuming an RBE of \(1,\) how many rem and rads is this dose, and how many joules of energy does a 75 kg person receive in a year? (c) How many chest x rays like the one in part (a) would it take to deliver the same total amount of energy to a 75 kg person as she receives from natural radiation in a year at sea level, as described in part (b)?

Comparison of energy released per gram of fuel. (a) When gasoline is burned, it releases \(1.3 \times 10^{8}\) J per gallon \((3.788\) L) of energy. Given that the density of gasoline is \(737 \mathrm{kg} / \mathrm{m}^{3},\) express the quantity of energy released in \(\mathrm{J} / \mathrm{g}\) of fuel. (b) During fission, when a neutron is absorbed by a \(^{235} \mathrm{U}\) nucleus, about 200 \(\mathrm{MeV}\) of energy is released for each nucleus that undergoes fission. Express this quantity in \(\mathrm{J} / \mathrm{g}\) 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 \(\mathrm{MeV}\) 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 meas- ured rate of \(3.92 \times 10^{26} \mathrm{W}\) . If its mass of \(1.99 \times 10^{30} \mathrm{kg}\) 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.)

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 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 \(^{60} \mathrm{Co}\) sources in your inventory was manufactured on October \(6,2008 .\) It is now April \(6,2011 .\) Is the source still usable? The half-life of \(^{60} \mathrm{Co}\) is 5.271 years.

Show that the net result of the proton-proton fusion chain that occurs inside our sun can be summarized as $$6 \mathrm{p}^{+} \rightarrow_{2}^{4} \mathrm{He}+2 \mathrm{p}^{+}+2 \beta^{+}+2 \gamma+2 \nu_{\mathrm{e}}$$

A proton and an antiproton annihilate, producing two photons. Find the energy, frequency, and wavelength of each photon emitted (a) if the initial kinetic energies of the proton and antiproton are negligible and (b) if each particle has an initial kinetic energy of 830 \(\mathrm{MeV} .\)

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