Two radioactive substances A and B are present as, [A] $=$ 8[B] and $t_{1 / 2}$ of A $=$ 10 minutes; $t_{1 / 2}$ of B $=$ 20 minutes. If they start decaying at the same time then after how much time concentration of both would be same:

20 min

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40 min

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60 min

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200 min

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Solution

The correct option is C 60 min
For the radioactive elements which follow first order kinetics $t_{1 / 2}$ is the time in which the concentration becomes half of the initial concentration.
For example $X$ has initial concentration $[X]$ than $[X]_{1 / 2} [X] / 2_{1 / 2} [X] / 4$
So for substance A which has concentration $[A] = 8 [B]$
So it decay will like this $[A] = 8 [B] 10 m i n \to 4 [B] 10 m i n \to 2 [B] 10 m i n \to [B] 10 m i n \to [B] / 2 10 m i n \to [B] / 4 10 m i n \to [B] / 8.............$
Same for B $[B] 20 m i n \to [B] / 2 20 m i n \to [B] / 4 20 m i n \to [B] / 8.............$

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Two radioactive substances A and B are present as, [A] = 8[B] and t1/2 of A = 10 minutes; t1/2 of B = 20 minutes. If they start decaying at the same time then after how much time concentration of both would be same:

Two radioactive substances A and B are present as, [A] $=$ 8[B] and $t_{1 / 2}$ of A $=$ 10 minutes; $t_{1 / 2}$ of B $=$ 20 minutes. If they start decaying at the same time then after how much time concentration of both would be same: