Problem 64 - Chemistry The Science in Context

Question

In a treatment that decreases pain and reduces inflammation of the lining of the knee joint, a sample of dysprosium-165 with a radioactivity of 1100 counts per second was injected into the knee of a patient suffering from rheumatoid arthritis. After \(24 \mathrm{h}\), the radioactivity had dropped to 1.14 counts per second. Calculate the half-life of \(^{165} \mathrm{Dy}\)

Step-by-Step Solution

Answer

Question: Determine the half-life of dysprosium-165 (\(^{165}Dy\)) if its initial radioactivity is 1100 counts per second, and after 24 hours, its radioactivity drops to 1.14 counts per second. Answer: The half-life of \(^{165}Dy\) is approximately 251,905 seconds.

Step 1: Write down the given information

Given information: Initial radioactivity: \(A_i = 1100 \hspace{2mm} \text{counts per second}\) Final radioactivity: \(A_f = 1.14 \hspace{2mm} \text{counts per second}\) Time of radioactivity drop: \(t = 24 \hspace{2mm} \text{hours} = 24 \times 60 \times 60 \hspace{2mm} \text{seconds}\)

Step 2: Write down the radioactive decay equation

Radioactive decay equation: \(A_f = A_i \times (0.5)^{\frac{t}{T_{\frac{1}{2}}}}\)

Step 3: Substitute the given values into the equation

Substitute the initial radioactivity (\(A_i\)), final radioactivity (\(A_f\)), and time (\(t\)): \(1.14 = 1100 \cdot (0.5)^{\frac{24\times60\times60}{T_{\frac{1}{2}}}}\)

Step 4: Solve for the half-life (\(T_{\frac{1}{2}}\))

To solve for \(T_{\frac{1}{2}}\), we will follow these steps: 1. Divide both sides by \(1100\). 2. Take the natural logarithm of both sides. 3. Solve for \(T_{\frac{1}{2}}\). Dividing both sides by \(1100\): \(\frac{1.14}{1100} = (0.5)^{\frac{24\times60\times60}{T_{\frac{1}{2}}}}\) Taking the natural logarithm of both sides: \(\ln(\frac{1.14}{1100}) = \ln((0.5)^{\frac{24\times60\times60}{T_{\frac{1}{2}}}})\) Using logarithmic properties: \(\frac{\ln(\frac{1.14}{1100})}{\ln(0.5)} = \frac{24\times60\times60}{T_{\frac{1}{2}}}\) Solving for \(T_{\frac{1}{2}}\): \(T_{\frac{1}{2}} = \frac{24\times60\times60}{\frac{\ln(\frac{1.14}{1100})}{\ln(0.5)}}\) Calculate the value: \(T_{\frac{1}{2}} \approx 251905 \hspace{2mm} \text{seconds}\) Finally, the half-life of \(^{165}Dy\) is approximately \(251,905\) seconds.

Key Concepts

Half-Life Calculation

Understanding the concept of half-life is crucial when dealing with radioactive substances. The half-life of a substance, like dysprosium-165, is the time it takes for half of the material to undergo radioactive decay. In practical terms, it tells us how quickly the material loses its radioactivity over time.

To find the half-life, we typically use the equation of radioactive decay:

• Initial radioactivity (\(A_i\)) - the starting radioactivity level.
• Final radioactivity (\(A_f\)) - the reduced radioactivity level after a given time.
• Time (\(t\)) - the duration of time over which the decay happens.
• Half-life (\(T_{\frac{1}{2}}\)) - the value we're solving for.

The equation \(A_f = A_i \times (0.5)^{\frac{t}{T_{\frac{1}{2}}}}\) helps us determine the half-life by measuring how much the substance decays in a known time frame. By rearranging the mathematical expressions and solving the equation, we determine the half-life of a radioactive substance. In this problem, by using values like initial and final counts per second, and the elapsed time of 24 hours, we've calculated the half-life of dysprosium-165 to be approximately 251,905 seconds.

Dysprosium-165

Dysprosium-165 is an isotope that plays an important role in medical treatments, especially for conditions like rheumatoid arthritis. This isotope is a radioactive form of the element dysprosium and is used in its decaying process to help treat medical conditions. Dysprosium itself is a soft metal known for its high melting point and other notable characteristics, mostly used in the technology sector.

However, its isotope, dysprosium-165, has found significant use in medicine. When injected into a joint affected by inflammation, the radioactivity of dysprosium-165 helps reduce the pain by decreasing inflammation. As it decays, it releases energy that disrupts the cellular processes of specific disease agents, reducing inflammation. Understanding its radioactive properties, including its half-life, helps in calculating the correct dosages and determining how long the treatment will be effective.

Rheumatoid Arthritis Treatment

Rheumatoid arthritis is an autoimmune condition where the body's immune system attacks its own tissues, leading to painful inflammation of the joints. Managing this condition involves reducing inflammation to alleviate pain and improve joint function.

One innovative approach is the use of isotopes like dysprosium-165. When injected directly into the joint, the radioactivity from dysprosium-165 disrupts the cells causing inflammation, offering relief from the swelling and pain associated with rheumatoid arthritis. This technique, referred to as radiosynoviorthesis, combines the precision of radioactivity with traditional anti-inflammatory treatment methods.

• Improves joint function by targeting inflamed tissues.
• Offers a localized treatment, which minimizes damage to healthy tissues.
• Can be more efficient over longer periods, depending on the isotope's half-life.

By comprehending the decay rate of dysprosium-165, medical professionals can better plan treatments according to individual needs, enhancing the quality of care for those suffering from rheumatoid arthritis.