Uncategorized: Practice Problem & Solution
50. In a radioactive material the activity at time $t_1$ is $R_1$ and at a later time $t_2$, it is $R_2$. If the decay constant of the material is $\lambda$, then: (2006)
Solution Explained:
To solve this problem, we apply the core principles of Uncategorized. Understanding the underlying formula is key to arriving at the correct answer below:
Activity follows the decay law $R = R_0 e^{-\lambda t}$. At $t_1$, $R_1 = R_0 e^{-\lambda t_1}$. At $t_2$, $R_2 = R_0 e^{-\lambda t_2}$. Dividing the two gives $R_1 / R_2 = e^{-\lambda (t_1 - t_2)}$, which rearranges to $R_1 = R_2 e^{-\lambda(t_1-t_2)}$.
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