Problem 22 - College Physics

Question

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.

Step-by-Step Solution

Answer

Yes, the source is still usable as it has an activity of 3615 Ci.

Step 1: Determine the Time Elapsed

Calculate the number of years between the original manufacturing date and the current date. The original date is October 6, 2008, and the current date is April 6, 2011. The time elapsed is 2 years and 6 months, which is equivalent to 2.5 years.

Step 2: Apply the Decay Formula

Use the exponential decay formula to calculate the remaining activity of the isotope after 2.5 years. The formula is \( A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \), where \( A_0 = 5000 \) Ci (initial activity), \( t = 2.5 \) years, and \( T_{1/2} = 5.271 \) years (half-life).

Step 3: Compute the Remaining Activity

Substitute the values into the formula: \[ A = 5000 \times \left( \frac{1}{2} \right)^{\frac{2.5}{5.271}} \]. Calculate \( \frac{2.5}{5.271} \approx 0.474 \) and then compute the power: \( 0.5^{0.474} \approx 0.723 \). Therefore, \( A \approx 5000 \times 0.723 = 3615 \) Ci.

Step 4: Determine Usability of the Source

The calculated activity is 3615 Ci. Compare this with the minimum usable activity of 3500 Ci to determine if the source is still usable.

Step 5: Conclusion

Since 3615 Ci is greater than 3500 Ci, the source is still usable for treatment.

Key Concepts

half-life calculations

Half-life calculations help us understand how quickly a radioactive substance decays. The half-life is the time taken for half of the radioactive atoms in a sample to decay. In our case, the half-life of cobalt-60 ( ^{60}Co) is 5.271 years. To find out if a radioactive source like ^{60}Co is still effective, we need to compute how much activity remains over time.

Here's how you can tackle half-life calculations:

• Determine elapsed time: Calculate the time from when the isotope was produced to the present day. In our scenario, we find it's 2.5 years since production.
• Apply the formula: Use the formula for exponential decay to find remaining activity.
• Compare with the threshold: Check if the remaining activity is above the needed threshold for medical applications, which is 3500 Ci in this case.

With this, you can conclude whether the radioisotope is still viable for cancer therapy.

radioisotopes in medicine

Radioisotopes play a crucial role in medical treatments and diagnostics. They are used mainly in cancer treatment, where their radioactivity can target and destroy cancer cells. Cobalt-60 ( ^{60}Co) is a commonly used radioisotope for this purpose.

Why are they used in medicine?

• Targeted therapy: Radioisotopes release radiation that can precisely target diseased cells.
• Diagnostic imaging: Radioisotopes help in diagnostic imaging, enabling doctors to visualize internal organs and structures.

In treatments, the effectiveness of radioisotopes depends on their activity level. If the activity falls below a certain level, the isotope can't generate enough radiation to be effective. That is why understanding and calculating half-life is so important in medicine. It helps ensure that the radioisotope is still potent enough for use in treatments.

exponential decay formula

The exponential decay formula is essential in determining how the activity of a radioactive material decreases over time. This formula is expressed as:\[ A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]where:

• A_0 is the original activity, in our case 5000 Ci.
• t is the time elapsed, which we've calculated as 2.5 years.
• T_{1/2} is the half-life, which is 5.271 years for ^{60}Co.

Using this formula helps us understand how quickly a radioisotope like ^{60}Co loses its radioactivity. With our calculation, we could see the activity decreased to 3615 Ci after 2.5 years. This is above the 3500 Ci threshold required for it to still be usable in medical treatments. Understanding this formula is vital for making predictions and ensuring safe and effective use of radioisotopes in various applications.