Problem 99 - Chemistry: The Central Science in SI Units, Global Edition

Question

Cobalt-60 is used in radiation therapy to treat cancer. It has a first-order rate constant for radioactive decay of \(k=1.31 \times 10^{-1} \mathrm{yr}^{-1}\). Another radioactive isotope, iron59, which is used as a tracer in the study of iron metabolism, has a rate constant of \(k=1.55 \times 10^{-2}\) day \(^{-1}\). (a) What are the half-lives of these two isotopes? (b) Which one decays at a faster rate? (c) How much of a 1.00-mg sample of each isotope remains after three half-lives? How much of a \(1.00-\mathrm{mg}\) sample of each isotope remains after five days?

Step-by-Step Solution

Answer

\(t_{1/2(Co-60)} \approx 5.27\,yr\), \(t_{1/2(Fe-59)} \approx 44.8\,days\); Iron-59 decays faster; After 3 half-lives: \(mass_{Co-60} \approx 0.125\,mg\), \(mass_{Fe-59} \approx 0.125\,mg\); After 5 days: \(mass_{Co-60(5 days)} \approx 0.959\,mg\), \(mass_{Fe-59(5 days)} \approx 0.597\,mg\).

Step 1: Calculate the half-life for Cobalt-60 and Iron-59

The half-life for a first-order reaction can be calculated using the formula: \[t_{1/2} = \frac{ln(2)}{k}\] For Cobalt-60, \(k = 1.31 \times 10^{-1}\,yr^{-1}\). Thus, \[t_{1/2(Co-60)} = \frac{ln(2)}{1.31 \times 10^{-1} yr^{-1}}\] For Iron-59, \(k = 1.55 \times 10^{-2}\,day^{-1}\). Thus, \[t_{1/2(Fe-59)} = \frac{ln(2)}{1.55 \times 10^{-2} day^{-1}}\]

Step 2: Determine the isotope that decays faster

Compare the rate constants to determine which isotope has the faster decay rate. Cobalt-60: \(k_{Co-60} = 1.31 \times 10^{-1} yr^{-1}\) Iron-59: \(k_{Fe-59} = 1.55 \times 10^{-2} day^{-1}\)

Step 3: Calculate the mass remaining after 3 half-lives

The remaining mass can be calculated using the half-life and the number of half-lives passed by using the formula: \[mass = initial \times \left(\frac{1}{2}\right)^n\] For both isotopes, we have an initial mass of \(1.00~mg\) and \(n = 3\). Cobalt-60: \[mass_{Co-60} = 1.00~mg \times \left(\frac{1}{2}\right)^3\] Iron-59: \[mass_{Fe-59} = 1.00~mg \times \left(\frac{1}{2}\right)^3\]

Step 4: Calculate the mass remaining after 5 days

Use the decay rate and the time duration to calculate the mass remaining, using the formula: \[mass = initial \times e^{-kt}\] For 5 days, we need the decay constant to be in days: Cobalt-60: Convert the rate constant to days using the conversion factor \(1 yr = 365 days\): \[k_{Co-60(day)} = 1.31 \times 10^{-1} yr^{-1} \times \frac{1}{365}\] Calculate the mass remaining after 5 days: \[mass_{Co-60(5 days)} = 1.00~mg \times e^{-k_{Co-60(day)} \times 5}\] Iron-59: Calculate the mass remaining after 5 days: \[mass_{Fe-59(5 days)} = 1.00~mg \times e^{-1.55 \times 10^{-2} day^{-1} \times 5}\]

Key Concepts

First-Order Reaction

In chemistry, a first-order reaction is a type where the rate of reaction depends linearly on the concentration of a single reactant. Such reactions are characterized by having a constant half-life regardless of the concentration of the substance. This means that for a first-order reaction, the time required for the reactant concentration to decrease by half is constant. Radioactive decay is a common example of a first-order reaction.
First-order reactions can be mathematically described by the equation:\[C_t = C_0 e^{-kt}\]Where:

• \( C_t \) is the concentration at time \( t \)
• \( C_0 \) is the initial concentration
• \( k \) is the rate constant
• \( t \) is time

This equation helps to determine how much of a substance remains after a certain period. Using the natural logarithm base \( e \) is pivotal in simplifying the calculations in the exponential decay formula.

Half-Life Calculation

The half-life of a substance is the time it takes for half of the sample to decay. It's a crucial aspect of understanding how long radioactive materials remain active and is particularly important in radioactive decay scenarios like those encountered with isotopes. For first-order reactions, the half-life is calculated using:\[t_{1/2} = \frac{\ln(2)}{k}\]Where:

• \( t_{1/2} \) is the half-life
• \( \ln(2) \) is approximately 0.693
• \( k \) is the rate constant

This calculation shows how long it takes for half of a radioactive sample to undergo decay. It is noteworthy that the half-life is independent of how much substance you start with, making it a convenient factor for comparing how quickly different substances decay.

Radiation Therapy

Radiation therapy is a medical treatment that uses high doses of radiation to kill cancer cells or shrink tumors. Cobalt-60 is one such radioactive substance used in this therapy. It emits gamma rays that are potent enough to damage the DNA of cancer cells, thereby preventing their reproduction or eliminating them.
In radiation therapy, accuracy is crucial to maximize the destruction of cancerous cells while minimizing damage to healthy tissue. This treatment might use external beam radiation or implants to deliver radiation directly to the tumor.
The inherent nature of radioactive decay in isotopes like Cobalt-60 is pivotal as it defines the therapy's duration and dosage, guided by its known half-life. This ensures the treatment is both effective and safe.

Tracer Isotope

Tracer isotopes play a significant role in medical and scientific research. Iron-59 is a well-known tracer used to study iron metabolism in biological systems. Tracers are isotopes that are added to a system to track the movement or concentration of elements or compounds within that system.
The utilization of Iron-59 enables researchers to understand how iron is absorbed, distributed, and processed by the body. This is highly beneficial for diagnosing and researching diseases related to iron metabolism, such as anemia or hemochromatosis.
Due to the radioactive nature of Iron-59, any changes in its behavior can be tracked by detecting the radiation it emits. This makes tracer isotopes incredibly valuable in medical imaging and functional studies, allowing scientists and doctors to obtain crucial data without invasive procedures.