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In the context of radioactive decay, the decay constant is a vital parameter that characterizes the rate at which a radioactive species, or isotope, decays. It is denoted by the Greek letter \( \lambda \), and typically has units of inverse time, such as \( \text{s}^{-1} \). The larger the decay constant, the faster the species decays. For example, in the original problem, species \( A \) has a decay constant of \( \lambda_{1}=1 \text{ sec}^{-1} \) and species \( B \) has \( \lambda_{2}=100 \text{ sec}^{-1} \). This means that species \( B \) decays much faster than species \( A \).
This is crucial because it affects the duration for which a species will be present in a particular amount, and when it will reach certain thresholds like maximum concentration.

Radioactive decay often involves the transformation of one species into another. This transformation can be understood through the decay equations showing how species \( A \) and \( B \) change over time. For example, species \( A \) decays into two units of \( B \), and then \( B \) further decays into three units of \( C \).
To understand this, consider radioactive decay as a chain reaction where:

• Species \( A \) initiates the transformation by breaking down into species \( B \), releasing energy and particles in the process.
• Species \( B \), once formed, quickly breaks down into species \( C \), again releasing energy and particles.

Thus, species transformation is a sequential process driven by inherent decay constants and the reaction properties of each species. It's a step-by-step breakdown from parent material to ultimately stable or non-radioactive end products.

Calculating moles in radioactive decay involves integrating the decay law to capture how the number of moles of a species changes over time. The fundamental expression \( N(t) = N_0 e^{-\lambda t} \) is used where \( N_0 \) is the initial number of moles, and the exponentiated term gives the decay over time.
For species \( B \), you integrate over time because it is formed from species \( A \). Therefore, its calculation accounts for both the appearing moles from \( A \)'s decay and its own decay. The integral:\[ N_B(t) = 2 \int_{0}^{t} N_A(t') \lambda_1 e^{-\lambda_2 (t - t')} dt' \]carves out the number of \( B \) moles at any time \( t \). This captures both the formation rate from \( A \) and the decay from \( B \).
In the given exercise, finding the maximum number of moles of \( B \) involves substituting the particular time, \( t_0 = 2 \ln(10) \), into the evaluated function of \( N_B(t) \), simplifying it with an approximation, and recognizing extremely small exponents tend towards zero. This results in the maximum number being \( \frac{200}{99} \) moles.