Nucleus: Practice Problem & Solution
A mixture consists of two radioactive materials $A_1$ and $A_2$ with half lives of 20 s and 10 s respectively. Initially the mixture has 40 g of $A_1$ and 160 g of $A_2$. The amount of the two in the mixture will become equal after: (2012 Pre)
Solution Explained:
To solve this problem, we apply the core principles of Nucleus. Understanding the underlying formula is key to arriving at the correct answer below:
Amounts remaining are equal: $40(1/2)^{t/20} = 160(1/2)^{t/10}$. This gives $(1/2)^{t/20 - t/10} = 4$, which means $2^{t/20} = 4 = 2^2$. Thus, $t/20 = 2$, yielding $t = 40 s$.
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