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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
A
40 s
B
60 s
C
80 s
D
20 s
Detailed Solution
$A_1$ (half-life 20 s): 40 g → 20 g (20 s) → 10 g (40 s)
$A_2$ (half-life 10 s): 160 g → 80 g (10 s) → 40 g (20 s) → 20 g (30 s) → 10 g (40 s)
Both become 10 g at t = 40 s.
Check: $40\left(\frac{1}{2}\right)^{t/20} = 160\left(\frac{1}{2}\right)^{t/10} \Rightarrow 2^{t/20} = 4 \Rightarrow t = 40$ s
$A_2$ (half-life 10 s): 160 g → 80 g (10 s) → 40 g (20 s) → 20 g (30 s) → 10 g (40 s)
Both become 10 g at t = 40 s.
Check: $40\left(\frac{1}{2}\right)^{t/20} = 160\left(\frac{1}{2}\right)^{t/10} \Rightarrow 2^{t/20} = 4 \Rightarrow t = 40$ s