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# A mixture consists of two radioactive materials A₁ and A₂ with half lives of 20s

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A mixture consists of two radioactive materials A₁ and A₂ with half lives of 20s and 10s respectively. Initially the mixture has 40 g of A₁ and 160 g of A₂. The amount of the two in the mixture will become equal after:

(a) 60 s
(b) 80 s
(c) 20 s
(d) 40 s

40 s

### Explanation:

Let, the amount of the two in the mixture will become equal after t years.
The amount of A₁, which remains after t years N₁ = N₀₁ / (2)ᵗ/²⁰
The amount of A₂, which remains after t years N₂ = N₀₂ / (2)ᵗ/¹⁰
According to the problem N₁ = N₂
40 / (2)ᵗ/²⁰ = 160 / (2)ᵗ/¹⁰ ⇒ 2ᵗ/²⁰ = 2(ᵗ/¹⁰⁻²)
t/20 = t/10 – 2 ⇒ t/20 – t/10 = 2
t/20 = 2 ⇒ t = 40 s

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