Problem 5 - Chemistry: The Central Science

Question

The half-life of cobalt-60 is \(5.27\) yr. How much of a \(1.000-\mathrm{mg}\) sample of cobalt-60 is left after \(15.81 \mathrm{yr}\) ? SOLUTION Analyze We are given the half-life for cobalt-60 and asked to calculate the amount of cobalt-60 remaining from an initial \(1.000\)-mg sample after \(15.81\) yr. Plan We will use the fact that the amount of a radioactive substance decreases by \(50 \%\) for every half-life that passes. Solve Because \(5.27 \times 3=15.81,15.81\) yr is three half-lives for cobalt-60. At the end of one half-life, \(0.500 \mathrm{mg}\) of cobalt-60 remains, \(0.250 \mathrm{mg}\) at the end of two half-lives, and \(0.125 \mathrm{mg}\) at the end of three half-lives.

Step-by-Step Solution

Answer

After 15.81 years (3 half-lives), the remaining amount of the cobalt-60 sample is \(0.125\ \mathrm{mg}\).

Step 1: Identify the number of half-lives

As given, the half-life of cobalt-60 is 5.27 years. To calculate the amount of cobalt-60 remaining after 15.81 years, we first need to determine how many half-lives have passed. Since \(15.81 \div 5.27 = 3\), 3 half-lives have passed.

Step 2: Calculate the remaining amount after the first half-life

After one half-life, the remaining amount of cobalt-60 would have decreased by 50%. So for our initial 1.000 mg sample, the remaining amount after the first half-life would be 50% of the initial amount, i.e., \(1.000 \times 0.5 = 0.500\ \mathrm{mg}\).

Step 3: Calculate the remaining amount after the second half-life

After the second half-life, the remaining amount will again decrease by 50%. This time, we will find 50% of the remaining amount after the first half-life (0.500 mg). So, \(0.500 \times 0.5 = 0.250\ \mathrm{mg}\) of cobalt-60 will be left after the second half-life.

Step 4: Calculate the remaining amount after the third half-life

After the third half-life, the remaining amount will once again decrease by 50%. We take 50% of the amount left after the second half-life (0.250 mg). This means the remaining amount of cobalt-60 would be \(0.250 \times 0.5 = 0.125\ \mathrm{mg}\).

Step 5: Conclusion

After 15.81 years (3 half-lives), the remaining amount of the cobalt-60 sample is 0.125 mg.

Key Concepts

Cobalt-60

Cobalt-60 is a synthetic radioactive isotope of cobalt with a mass number of 60. This means it has 27 protons and 33 neutrons in its nucleus. It is produced artificially in nuclear reactors through the neutron activation of cobalt-59.

Cobalt-60 has a wide range of applications, including use in cancer therapy as a radiation source for radiotherapy, sterilization of medical equipment, and as a tracer element in various industries. Its half-life, the time it takes for half of a given amount of cobalt-60 to decay, is approximately 5.27 years. Understanding how cobalt-60 decays over time is critical for these applications, as it ensures correct dosing in medical treatments and safe handling across various uses.

One important aspect in handling cobalt-60 is knowing how much of it remains active after a certain period, which often involves calculations using its half-life. Such calculations are based on the principles of exponential decay, a concept underlying the decrease in the quantity of this substance over time.

Radioactive Decay

Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation in the form of particles or electromagnetic waves. This spontaneous transformation leads to the formation of a different element or a different isotope of the same element. Radioactive decay is a random process at the level of single atoms, meaning that it is impossible to predict exactly which atom will decay at a given moment.

There are several types of radioactive decay, including alpha decay, beta decay, and gamma decay. Cobalt-60 undergoes beta decay, where a neutron in the nucleus is transformed into a proton and an electron which is then ejected from the atom. This process also releases a gamma photon, which is a form of high-energy electromagnetic radiation. The rate of decay is characterized by the half-life, which varies between different isotopes. For cobalt-60, with each half-life that passes, the remaining quantity of the substance decreases by half - demonstrating a classic exponential decay pattern.

Exponential Decay

Exponential decay is a distinctive decrease over time, where the quantity of a decaying substance reduces at a rate proportional to its current amount. This type of decay can be represented mathematically with the formula:

\[ N(t) = N_0 e^{-\textstyle{\frac{t}{\tau}}} \]
Where:\

• \(N(t)\) is the quantity that still remains and has not yet decayed after time \(t\),
• \(N_0\) is the initial quantity of the substance,
• \(e\) is the base of natural logarithms,
• and \(\tau\) is the mean lifetime of the process.

In particular, when considering the half-life \(t_{1/2}\), the formula can be simplified since the substance is halved each half-life, leading to:\
\[ N(t) = N_0 \times (\frac{1}{2})^{\frac{t}{t_{1/2}}} \]
For instance, if cobalt-60 has a half-life of 5.27 years, after one half-life, only 50% of its original amount remains. This process then repeats itself each half-life, with the amount present decreasing by half from its previous amount, resulting in a pattern where the quantity of the substance diminishes rapidly initially, then more slowly over time.

This concept explains the solution process for determining how much cobalt-60 remains after a certain number of years. To comprehend these calculations, one must fully grasp how the decay rate, expressed through the half-life, impacts the amount of radioactive material over time.