Two radioactive elements R and S disintegrates as
$R \to P + α; λ_{R} = 4.5 \times 10^{- 3} y e a r s^{- 1} S \to P + β; λ_{S} = 3 \times 10^{- 3} y e a r s^{- 1}$
Starting with number of atoms of R and S in the ratio of 2 : 1, this ratio after the lapse of three half lives of R will be :

3 : 2

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1 : 3

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1 : 1

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2 : 1

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Solution

The correct option is C 1 : 1
$N_{1} = N_{01} e^{- λ_{1} t} N_{2} = N_{02} e^{- λ_{2} t} \frac{N_{1}}{N_{2}} = \frac{2}{1} \times e^{- (λ_{1} - λ_{2})} t λ_{1} t = 3 l n 2 λ_{2} t = 3 l n 2 \times \frac{3}{4.5} = 2 l n 2 \Rightarrow \frac{N_{1}}{N_{2}} = 2 \times 3^{- l n 2} = 1$

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Two radioactive elements R and S disintegrates as R→P+α;λR=4.5×10−3years−1S→P+β;λS=3×10−3years−1 Starting with number of atoms of R and S in the ratio of 2 : 1, this ratio after the lapse of three half lives of R will be :

The correct option is C 1 : 1N1=N01e−λ1tN2=N02e−λ2tN1N2=21×e−(λ1−λ2)tλ1t=3ln2λ2t=3ln2×34.5=2ln2⇒N1N2=2×3−ln2=1

Two radioactive elements R and S disintegrates as
$R \to P + α; λ_{R} = 4.5 \times 10^{- 3} y e a r s^{- 1} S \to P + β; λ_{S} = 3 \times 10^{- 3} y e a r s^{- 1}$
Starting with number of atoms of R and S in the ratio of 2 : 1, this ratio after the lapse of three half lives of R will be :

The correct option is C 1 : 1
$N_{1} = N_{01} e^{- λ_{1} t} N_{2} = N_{02} e^{- λ_{2} t} \frac{N_{1}}{N_{2}} = \frac{2}{1} \times e^{- (λ_{1} - λ_{2})} t λ_{1} t = 3 l n 2 λ_{2} t = 3 l n 2 \times \frac{3}{4.5} = 2 l n 2 \Rightarrow \frac{N_{1}}{N_{2}} = 2 \times 3^{- l n 2} = 1$