A mixture consists of two radioactive materials A1 and

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A mixture consists of two radioactive materials A₁ and A₂ with half lives of 20 s and 10 s 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 :

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	=	160
	(2)^t/20		(2)^t/10

2^t/20 = 2^{(t/10) - 2}

	t	=	t	- 2
	20		10

	t	-	t	= 2
	20		10

	t	= 2
	200

t = 40 s