Problem 52
Question
(II) A sample of \(\frac{233}{92} \mathrm{U} \quad\left(T_{2}=1.59 \times 10^{5} \mathrm{yr}\right)\) contains \(5.50 \times 10^{18}\) nuclei. \((a)\) What is the decay constant? \((b)\) Approximately how many disintegrations will occur per minute?
Step-by-Step Solution
Verified Answer
The decay constant is approximately \(8.287 \times 10^{-12} \text{ min}^{-1}\). There are about \(4.558 \times 10^{7}\) disintegrations per minute.
1Step 1: Calculate the Decay Constant
The decay constant \( \lambda \) is related to the half-life \( T_{1/2} \) by the formula:\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]Given, the half-life \( T_{1/2} = 1.59 \times 10^{5} \text{ years} \). First, convert the half-life from years to minutes:\[ 1 \text{ year} = 525600 \text{ minutes} \] So,\[ T_{1/2} = 1.59 \times 10^{5} \text{ years} \times 525600 \text{ minutes/year} = 8.35844 \times 10^{10} \text{ minutes} \]Now, calculate the decay constant:\[ \lambda = \frac{\ln(2)}{8.35844 \times 10^{10}} \approx 8.287 \times 10^{-12} \text{ min}^{-1} \]
2Step 2: Calculate the Activity
The activity \( A \) of a radioactive sample, which is the number of disintegrations per time, is given by:\[ A = \lambda N \]where \( N = 5.50 \times 10^{18} \) is the number of nuclei.Substitute the known values:\[ A = 8.287 \times 10^{-12} \text{ min}^{-1} \times 5.50 \times 10^{18} = 4.558 \times 10^{7} \text{ disintegrations per minute} \]
Key Concepts
Decay ConstantHalf-Life CalculationsNuclear Disintegration
Decay Constant
The decay constant, denoted by \( \lambda \), is a fundamental property of a radioactive substance. It provides the probability per unit time that a single nucleus will decay.
The decay constant is inversely related to the half-life of the substance, which is the time required for half of the radioactive nuclei to disintegrate. The formula to calculate the decay constant is:
In practical terms, understanding the decay constant helps us compute how quickly or slowly a sample will decay, thus influencing areas such as nuclear medicine and radiometric dating. For example, a large decay constant indicates rapid decay, while a small one suggests slower decay over a longer period.
The decay constant is inversely related to the half-life of the substance, which is the time required for half of the radioactive nuclei to disintegrate. The formula to calculate the decay constant is:
- \( \lambda = \frac{\ln(2)}{T_{1/2}} \)
In practical terms, understanding the decay constant helps us compute how quickly or slowly a sample will decay, thus influencing areas such as nuclear medicine and radiometric dating. For example, a large decay constant indicates rapid decay, while a small one suggests slower decay over a longer period.
Half-Life Calculations
Half-life calculations form a critical part of understanding radioactive decay. The half-life of a radioactive substance is the time it takes for half of the radioactive nuclei in a sample to decay. This property is essential as it helps to predict how the concentration of a radioisotope decreases over time.
To perform half-life calculations, consider the equation:
Understanding half-life assists both in practical applications and theoretical calculations, enabling scientists to design processes or predict changes appropriately.
To perform half-life calculations, consider the equation:
- \( T_{1/2} = \frac{\ln(2)}{\lambda} \)
Understanding half-life assists both in practical applications and theoretical calculations, enabling scientists to design processes or predict changes appropriately.
Nuclear Disintegration
Nuclear disintegration, commonly referred to as decay, is the process by which an unstable atomic nucleus loses energy by emitting radiation. This process leads to the transformation of the original nucleus into one or more different elements or isotopes.
In the context of radioactive decay, the activity (or rate of disintegration) is expressed as the number of disintegrations that occur per unit of time. This is often measured in units like disintegrations per minute. The formula used to calculate this activity is:
Essentially, the activity provides insight into how active a radioactive sample is at a given time and how quickly it will reach stability. For students and professionals, understanding nuclear disintegration is key for fields that rely on radiation, such as nuclear medicine, energy, and archaeology.
In the context of radioactive decay, the activity (or rate of disintegration) is expressed as the number of disintegrations that occur per unit of time. This is often measured in units like disintegrations per minute. The formula used to calculate this activity is:
- \( A = \lambda N \)
Essentially, the activity provides insight into how active a radioactive sample is at a given time and how quickly it will reach stability. For students and professionals, understanding nuclear disintegration is key for fields that rely on radiation, such as nuclear medicine, energy, and archaeology.
Other exercises in this chapter
Problem 50
(II) Calculate the activity of a pure \(8.7-\mu \mathrm{g}\) sample of \({ }_{15}^{32} \mathrm{P}\left(T_{\frac{1}{2}}=1.23 \times 10^{6} \mathrm{~s}\right)\)
View solution Problem 52
(II) A sample of \({ }_{92}^{233} \mathrm{U} \quad\left(T_{1}=1.59 \times 10^{5} \mathrm{yr}\right)\) contains \(5.50 \times 10^{18}\) nuclei. \(\quad(a)\) What
View solution Problem 53
(II) The activity of a sample drops by a factor of 4.0 in 8.6 minutes. What is its half-life?
View solution Problem 56
(II) Rubidium-strontium dating. The rubidium isotope \({ }_{37}^{87} \mathrm{Rb}\), a \(\beta\) emitter with a half-life of \(4.75 \times 10^{10} \mathrm{yr}\),
View solution