Assuming that many radioactive nuclides can be considered safe after 20 half- lives, how long will it take for each of the following nuclides to be safe: (a) \({ }^{242} \mathrm{Cm}\left(t_{1 / 2}=163\right.\) days \() ;\) (b) \({ }^{214} \mathrm{Po}\) \(\left(t_{1 / 2}=1.6 \times 10^{-4} \mathrm{~s}\right) ;(\mathrm{c})^{232} \mathrm{Th}\left(t_{1 / 2}=1.39 \times 10^{10} \mathrm{yr}\right) ?\)

Short Answer

Expert verified
For \({ }^{242} \mathrm{Cm}\): 3260 days; \({ }^{214} \mathrm{Po}\): 3.2 \ \times 10^{-3} \ \text{s}; \({ }^{232} \mathrm{Th}\): 2.78 \ \times 10^{11} \ \text{yr}\).

Step by step solution

01

- Understand the Half-Life Formula

The half-life of a nuclide is the time it takes for half of the radioactive atoms in a sample to decay. We need to find the time it takes for a nuclide to become safe, which is considered to be after 20 half-lives.
02

- Calculate the Time for \({ }^{242} \mathrm{Cm}\)

For \({ }^{242} \mathrm{Cm}\), the half-life \(t_{1/2} \) is 163 days. To find the time for the nuclide to be safe: \( T = 20 \times t_{1/2} = 20 \times 163 \ \text{days} = 3260 \ \text{days} \).
03

- Calculate the Time for \({ }^{214} \mathrm{Po}\)

For \({ }^{214} \mathrm{Po}\), the half-life \(t_{1/2} \) is \(1.6 \times 10^{-4} \ \text{s}\). The time to be safe is: \( T = 20 \times t_{1/2} = 20 \times 1.6 \times 10^{-4} \ \text{s} = 3.2 \ \times 10^{-3} \ \text{s} \).
04

- Calculate the Time for \({ }^{232} \mathrm{Th}\)

For \({ }^{232} \mathrm{Th}\), the half-life \(t_{1/2} \) is \(1.39 \times 10^{10} \ \text{yr}\). The time to be safe is: \( T = 20 \times t_{1/2} = 20 \times 1.39 \times 10^{10} \ \text{yr} = 2.78 \ \times 10^{11} \ \text{yr} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radioactive Decay
Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. This can occur through various types of decay, such as alpha, beta, and gamma decay. The key point to understand is that radioactive decay is a random process; however, statistical methods allow us to predict how a large quantity of identical nuclides will behave over time. The rate at which a nuclide decays is characterized by its half-life.
Half-Life Formula
The half-life of a radioactive substance is the time required for half of any given amount of the substance to decay. The general formula can be represented as: \[ T = n \times t_{1/2} \]where \( T \) is the total time taken for the nuclide to become safe, \( n \) is the number of half-lives, and \( t_{1/2} \) is the half-life of the substance. In most safety assessments, a material is considered safe after 20 half-lives. Given the half-life, calculating the time to be safe is straightforward: just multiply the half-life by 20. This is what we did in the exercises for \(^{242} \text{Cm}\), \(^{214} \text{Po}\), and \(^{232} \text{Th}\).
Nuclide Safety Assessment
Nuclide safety assessment involves determining how long a radioactive material needs to be stored or managed before it poses minimal risk to humans and the environment. Safety is often assessed based on the concept of half-life. For instance, if a substance has a very short half-life, it will decay quickly and might pose a high initial risk but will become safe relatively soon. Conversely, substances with long half-lives, like \(^{232} \text{Th}\), remain radioactive for extended periods and require long-term management. In the example given, \(^{214} \text{Po} \left(t_{1/2}=1.6 \times 10^{-4} \text{~s}\right)\) becomes safe in just a matter of milliseconds, whereas \(^{232} \text{Th} \left(t_{1/2}=1.39 \times 10^{10} \text{yr}\right)\) takes billions of years to reach the same level of safety. Understanding the half-life helps in making informed decisions about handling and storing radioactive materials.

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