In a state of radioactive secular equilibrium, the atomic ratio of a highly stable parent element to — Nuclear Chemistry and Radioactivity Chemistry Question
Question
In a state of radioactive secular equilibrium, the atomic ratio of a highly stable parent element $A$ to its radioactive daughter element $B$ is found to be exactly $10^6 : 1$. If the half-life of the daughter element $B$ is exactly $5\text{ hours}$, calculate the half-life of the parent element $A$ in hours. Provide the value $x$ where $T_{1/2(A)} = x \times 10^6\text{ hours}$.
💡 Solution & Explanation
For a system in radioactive equilibrium, the decay rates are equal: $N_A \lambda_A = N_B \lambda_B$, which can be written as $\frac{N_A}{T_{1/2(A)}} = \frac{N_B}{T_{1/2(B)}}$. Rearranging this yields $T_{1/2(A)} = T_{1/2(B)} \times (\frac{N_A}{N_B})$. Substituting the given values: $T_{1/2(A)} = 5\text{ hours} \times (10^6 / 1) = 5 \times 10^6\text{ hours}$. Thus, $x = 5$.