Chapter 5: Problem 11
\({ }^{218} \mathrm{Po}\) is an \(\alpha\) emitter. How many \(\alpha\) particles are emitted each second from a milligram of pure \({ }^{218}\) Po whose half- life is \(3.05\) minutes?
Short Answer
Expert verified
Approximately \(1.05 \times 10^{16}\) \(\alpha\) particles are emitted per second.
Step by step solution
01
Determine the Decay Constant
The decay constant \( \lambda \) can be calculated using the formula \( \lambda = \frac{0.693}{T_{1/2}} \), where \( T_{1/2} \) is the half-life. Given that the half-life \( T_{1/2} \) of \( ^{218}\text{Po} \) is 3.05 minutes, or 183 seconds, we calculate \( \lambda \) as follows: \[ \lambda = \frac{0.693}{183 \text{ seconds}} = 0.00379 \text{ s}^{-1} \]
02
Calculate the Number of Atoms in 1 mg of \(^{218}\text{Po}\)
First, we find the molar mass of \(^{218}\text{Po}\), which is approximately 218 g/mol. Since 1 mg is \(1 \times 10^{-3}\) g, we calculate the number of moles as:\[ \text{moles} = \frac{1 \times 10^{-3} \text{ grams}}{218 \text{ g/mol}} = 4.59 \times 10^{-6} \text{ mol} \]Now, using Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol), the number of atoms \( N \) is:\[ N = 4.59 \times 10^{-6} \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} = 2.77 \times 10^{18} \text{ atoms} \]
03
Calculate the Activity (Number of \(\alpha\) Particles Emitted per Second)
The activity \( A \), which measures how many \(\alpha\) particles are emitted per second, is given by the formula \( A = \lambda N \). Using the \(\lambda\) from Step 1 and \(N\) from Step 2:\[ A = 0.00379 \text{ s}^{-1} \times 2.77 \times 10^{18} \text{ atoms} \]\[ A \approx 1.05 \times 10^{16} \text{ \(\alpha\) particles/second} \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polonium-218
Polonium-218 (\(^{218}\mathrm{Po}\)) is a radioactive isotope that plays a significant role in the concept of alpha decay. This isotope undergoes decay by emitting an alpha particle, which consists of 2 protons and 2 neutrons. This results in a decrease in the atomic number by two, transforming Polonium (\(\mathrm{Po}\)) into another element. Often, the product of this decay is Lead-214 (\(^{214}\mathrm{Pb}\)).
Some interesting characteristics about \(^{218}\mathrm{Po}\) include:
Some interesting characteristics about \(^{218}\mathrm{Po}\) include:
- It is a part of the Uranium decay series.
- It has a relatively short half-life of only 3.05 minutes.
- Its atomic mass is 218, indicating it contains a combination of protons and neutrons adding to this number.
Half-life Calculation
The half-life is a fundamental concept in understanding radioactive materials and decay processes. It is defined as the time required for half of a radioactive sample to decay. For Polonium-218, this duration is 3.05 minutes. Understanding and calculating half-life is vital in predicting how quickly a nuclear sample becomes less radioactive over time.
Here is how one would calculate the decay constant \( \lambda \), which is essential in determining activity:
Here is how one would calculate the decay constant \( \lambda \), which is essential in determining activity:
- Use the formula \( \lambda = \frac{0.693}{T_{1/2}} \) where \( T_{1/2} \) is the half-life duration.
- For \( ^{218}\mathrm{Po} \) with a half-life of 183 seconds (3.05 minutes), \( \lambda \approx 0.00379\,\mathrm{s}^{-1} \).
Radioactive Decay Activity
Radioactive decay activity measures the rate at which a radioactive sample emits particles, key to understanding radiative processes in substances like Polonium-218. This activity is expressed in terms of the number of decay events per second, commonly called the Becquerel (Bq) where 1 Bq = 1 decay per second.
For \( ^{218}\mathrm{Po} \), you would calculate activity as follows:
For \( ^{218}\mathrm{Po} \), you would calculate activity as follows:
- First, determine the number of atoms in a given sample using its molar mass and Avogadro's number.
- Then, multiply the decay constant \( \lambda \) by the number of atoms \( N \) to find the activity \( A \).
- Example: For a 1 mg sample of \( ^{218}\mathrm{Po} \) at 0.00379 s\(^{-1}\) decay constant and \( 2.77 \times 10^{18} \) atoms, \( A \approx 1.05 \times 10^{16} \) particles/second.