Question:
Two radioactive materials $A_1$ and $A_2$ have half-life periods 20 s and 10 s respectively. Initially a mixture of these materials contain 40 g of $A_1$ and 160 g of $A_2$. The time taken for $A_1$ and $A_2$ to become equal in the mixture is:
Two radioactive materials $A_1$ and $A_2$ have half-life periods 20 s and 10 s respectively. Initially a mixture of these materials contain 40 g of $A_1$ and 160 g of $A_2$. The time taken for $A_1$ and $A_2$ to become equal in the mixture is:
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Set up the ratio of remaining amounts to solve for time $t$.
Updated On: Jun 6, 2026
- 60 s
- 80 s
- 20 s
- 40 s
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Radioactive decay formula: $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$.
Step 2: Meaning
Find time $t$ such that $N_{A1}(t) = N_{A2}(t)$.
Step 3: Analysis
$N_{A1} = 40 (1/2)^{t/20}$, $N_{A2} = 160 (1/2)^{t/10}$. $40 (1/2)^{t/20} = 160 (1/2)^{t/10}$. $(1/2)^{t/20} / (1/2)^{t/10} = 160 / 40 = 4$. $(1/2)^{t/20 - 2t/20} = 4 \rightarrow (1/2)^{-t/20} = 4 \rightarrow 2^{t/20} = 2^2$. $t/20 = 2 \rightarrow t = 40$ seconds.
Step 4: Conclusion
The amounts become equal after 40 seconds.
Final Answer: (D)
Radioactive decay formula: $N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$.
Step 2: Meaning
Find time $t$ such that $N_{A1}(t) = N_{A2}(t)$.
Step 3: Analysis
$N_{A1} = 40 (1/2)^{t/20}$, $N_{A2} = 160 (1/2)^{t/10}$. $40 (1/2)^{t/20} = 160 (1/2)^{t/10}$. $(1/2)^{t/20} / (1/2)^{t/10} = 160 / 40 = 4$. $(1/2)^{t/20 - 2t/20} = 4 \rightarrow (1/2)^{-t/20} = 4 \rightarrow 2^{t/20} = 2^2$. $t/20 = 2 \rightarrow t = 40$ seconds.
Step 4: Conclusion
The amounts become equal after 40 seconds.
Final Answer: (D)
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