Question

Consider the decay series \(\mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C} \longrightarrow \mathrm{D}\) where \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are radioactive isotopes with halflives of \(4.50 \mathrm{~s}, 15.0\) days, and \(1.00 \mathrm{~s},\) respectively, and \(\mathrm{D}\) is nonradioactive. Starting with 1.00 mole of \(\mathrm{A},\) and none of \(\mathrm{B}, \mathrm{C},\) or \(\mathrm{D},\) calculate the number of moles of \(\mathrm{A}\), \(\mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) left after 30 days.

Short answer

After 30 days: 0 mol A, ~0.999 mol B, ~0 mol C, ~1 mol D.

Step-by-step solution

Calculate Decay Constants

The decay constant \( \lambda \) for a substance can be calculated using the formula: \( \lambda = \frac{\ln(2)}{t_{1/2}} \), where \( t_{1/2} \) is the half-life of the isotope. \( \lambda_A = \frac{\ln(2)}{4.50 \, \text{s}} \approx 0.154 \; \text{s}^{-1} \) for \( \mathrm{A} \). \( \lambda_B = \frac{\ln(2)}{15 \times 86400 \, \text{s}} \approx 5.34 \times 10^{-7} \; \text{s}^{-1} \) for \( \mathrm{B} \). \( \lambda_C = \frac{\ln(2)}{1.00 \, \text{s}} \approx 0.693 \; \text{s}^{-1} \) for \( \mathrm{C} \).

Determine Remaining A after 30 Days

The number of moles of a radioactive isotope remaining after time \( t \) is given by \( N(t) = N_0 e^{-\lambda t} \). For isotope \( \mathrm{A} \): \[ N_A(30 \, \text{days}) = 1.00 \, \text{mol} \times e^{-0.154 \, \text{s}^{-1} \times 30 \times 86400 \, \text{s}} \approx 0 \, \text{mol} \]. This approximation indicates that \( \mathrm{A} \) will have decayed significantly and essentially is completely decayed after 30 days.

Calculate Intermediate B Accumulation

Since \( \mathrm{B} \) decays slower than it accumulates, it almost reaches equilibrium. The growth function for \( \mathrm{B} \) is \( N_B = \frac{\lambda_A}{\lambda_B - \lambda_A} N_0 (e^{-\lambda_A t} - e^{-\lambda_B t}) \). For 30 days: \[ N_B(30 \, \text{days}) \approx \frac{0.154}{5.34 \times 10^{-7} - 0.154} 1.00 \times (0 - e^{-5.34 \times 10^{-7} \times 30 \times 86400}) \approx 0.999 \]. Thus, the number of \( \mathrm{B} \) is approximately 0.999 mol.

Calculate C Accumulation and Decay

Since \( \mathrm{C} \) has a very short half-life, the approximate number of \( \mathrm{C} \) molecules is negligible: \( N_C(t) \approx 0 \). This is because the fast decay of \( \mathrm{C} \) balances rapid formation from \( \mathrm{B} \), keeping \( \mathrm{C} \) close to zero mol.

Calculate D Accumulation

Since \( \mathrm{D} \) is a non-radioactive decay product from \( \mathrm{C} \), the number of moles for \( \mathrm{D} \) will equate to the initial moles minus the moles still in radioactive forms. So,\( N_D = 1.00 \text{ mol} - 0.0 \approx 1.00 \text{ mol} \). Thus, approximately 1.00 mol of \( \mathrm{D} \) formed as the final decay product after 30 days.

Key concepts

Half-life Calculation

The half-life of a radioactive substance is the time it takes for half of the original quantity of isotopes to decay. It is a crucial concept when analyzing radioactive decay series because it helps us understand how quickly a substance transforms. To calculate the half-life, we make use of the decay constant.
The relationship can be described using the formula:
\[ t_{1/2} = \frac{\ln(2)}{\lambda} \]
where \( \ln(2) \) is the natural logarithm of 2, approximately equal to 0.693. This formula emphasizes that the time it takes for half of the substance to decay is inversely proportional to the decay constant.
In the decay series we are studying, each isotope A, B, and C has its own half-life. For example, isotope A decays much more rapidly than isotope B because its half-life is just 4.5 seconds as compared to the 15-day half-life of isotope B.

Decay Constant

The decay constant, symbolized as \( \lambda \), is a parameter that quantifies the rate of decay of a radioactive isotope. It is a measure of the probability per unit time that a single atom will decay. A higher decay constant indicates a faster decay process.
To calculate the decay constant, you can use the formula:
\[ \lambda = \frac{\ln(2)}{t_{1/2}} \]
This formula highlights that the decay constant depends on the isotope's half-life, where \( \ln(2) \) is approximately 0.693. The decay constants for isotopes A, B, and C in our scenario were calculated based on their respective half-lives.

• For \( A \), \( \lambda_A = 0.154 \, \text{s}^{-1} \).
• For \( B \), \( \lambda_B = 5.34 \times 10^{-7} \, \text{s}^{-1} \).
• For \( C \), \( \lambda_C = 0.693 \, \text{s}^{-1} \).

By knowing the decay constants, you can easily predict how much of each isotope will remain after a certain period.

Radioactive Isotopes

Radioactive isotopes, or radioisotopes, are variants of chemical elements with unstable nuclei that release radiation as they decay to more stable forms. In a decay series, like our example, radioisotopes transform into one another step-by-step until a stable isotope is formed.
The decay series begins with isotope A, a radioactive isotope, that decays to B, another unstable isotope.

• Isotope A has a half-life of 4.50 seconds, indicating rapid decay.
• Isotope B, with a half-life of 15 days, decays more slowly.
• Isotope C has a half-life of 1.00 second, making it very transient.

This series ultimately ends with D, a nonradioactive and stable product. During decay, each radioactive isotope in the series will have its amount traceable and predictable based on its decay rate, calculated using the exponential decay formula.

Exponential Decay Formula

The exponential decay formula is crucial for calculating the remaining quantity of a radioactive isotope after a certain period. It mathematically expresses how radioactive isotopes decrease over time. This formula is:
\[ N(t) = N_0 e^{-\lambda t} \]
where:

• \( N(t) \) is the remaining amount after time \( t \)
• \( N_0 \) is the initial amount
• \( \lambda \) is the decay constant
• \( e \) is the base of the natural logarithm, approximately 2.718

In practical terms, by inserting values into this formula, you can determine how much of an isotope remains at a given point in time.
Considering that isotope A has a high decay constant, the formula shows why nearly none remains after 30 days. By understanding this concept, one can comprehend how quickly or slowly different isotopes undergo decay over time.