Problem 37 - Chemistry and Chemical Reactivity

Question

Radioactive cobalt-60 is used extensively in nuclear medicine as a \(\gamma\) ray source. It is made by a neutron capture reaction from cobalt-59 and is a \(\beta\) emitter; \beta emission is accompanied by strong \(\gamma\) radiation. The half-life of cobalt-60 is 5.27 years. (a) How long will it take for a cobali-60 source to decrease to one eighth of its original activity? (b) What fraction of the activity of a cobalt-60 source remains after 1.0 year?

Step-by-Step Solution

Answer

(a) 15.81 years; (b) 87.6% remaining after 1 year.

Step 1: Understand Half-Life Concept

Cobalt-60 undergoes radioactive decay, where its nucleus emits a beta particle. The half-life is the time it takes for half of a given amount of radioactive substance to decay. For example, after one half-life (5.27 years for Cobalt-60), half of the original Cobalt-60 atoms will have decayed, leaving half of the initial radioactive material.

Step 2: Use Half-Life to Find Time to Decrease Activity to One Eighth

Each half-life reduces the activity by half. To find the time it takes for the activity to decrease to one eighth, we need to find the number of half-lives until one eighth of the original activity remains: - First half-life: 1/2 of original - Second half-life: 1/4 of original - Third half-life: 1/8 of original Thus, after three half-lives, the activity will decrease to one eighth of the original activity. Since the half-life of Cobalt-60 is 5.27 years, the total time for three half-lives is \[3 \times 5.27 = 15.81\, \text{years}\]

Step 3: Fraction Remaining After 1 Year

Use the exponential decay formula to calculate the remaining fraction after 1 year. The formula for decay is \[N(t) = N_0 (1/2)^{t/T}\] where \(N(t)\) is the remaining activity, \(N_0\) is the initial activity, \(t\) is the elapsed time, and \(T\) is the half-life. Substitute \(t = 1\) year and \(T = 5.27\) years: \[N(1) = N_0 \left(\frac{1}{2}\right)^{1/5.27}\] Calculating this gives the fraction: \[N(1) = N_0 (0.876)\] Thus, approximately 87.6% of the original activity remains after 1 year.

Key Concepts

Half-Life

The concept of half-life is crucial to understanding radioactive decay. It represents the time required for half of a radioactive isotope, like Cobalt-60, to lose its activity and decay. When we say Cobalt-60 has a half-life of 5.27 years, it means every 5.27 years, half of any sample will have undergone decay. This consistent reduction makes the half-life a powerful tool in predicting how a sample's activity changes over time.

Here's how it works:

• After the first half-life, you have 1/2 of the original amount.
• After the second half-life, only 1/4 of the original remains (as you half again from the previous step).
• This pattern continues, halving each time, so after three half-lives, you are left with 1/8 of the original amount.

Understanding half-life allows us to determine how long it takes for a substance to reach a certain level of decay. In practical terms, this helps scientists and engineers plan the safe handling and disposal of radioactive materials.

Cobalt-60

Cobalt-60 ( ^{60} ext{Co} ) is a radioactive isotope of cobalt widely used in medicine and industry. It is produced when cobalt-59 ( ^{59} ext{Co} ) captures a neutron, creating Cobalt-60. This isotope is a strong beta emitter and is accompanied by gamma radiation, making it highly useful in various applications.

In nuclear medicine, Cobalt-60 is often used as a gamma-ray source for radiotherapy, effectively targeting cancer cells with precision. Its predictable decay and emission properties allow for targeted treatments.

However, its radioactivity means that careful planning and safety protocols are necessary. Handling requires trained professionals to minimize exposure, especially since its gamma radiation can penetrate the body. Its 5.27 year half-life also makes it suitable for long-term medical applications, but also obligates periodic evaluation and eventual disposal once it reduces in efficacy.

Exponential Decay Formula

The exponential decay formula is a mathematical model to calculate the reduction of radioactive material over time. It helps us to precisely understand how quickly a substance like Cobalt-60 decreases in activity. The formula is expressed as:\[N(t) = N_0 (1/2)^{t/T}\]where:

• \(N(t)\) is the remaining activity after time \(t\).
• \(N_0\) is the initial activity.
• \(t\) is the time that has passed.
• \(T\) is the half-life of the substance.

Using this formula, we can determine how much of the radioactive material remains after a specific period. For instance, if you want to know what fraction of Cobalt-60 remains after 1 year, you substitute \(t = 1\) and \(T = 5.27\), the half-life of Cobalt-60.The result is that approximately 87.6% of the original Cobalt-60 activity remains after 1 year. This specific and predictable mathematical approach is vital for practical applications, such as calculating safe exposure levels and determining when a radioactive source needs replacement in medical or industrial settings.