Half lives of two radioactive substances A and B are respectively 20 min and 40 min. Initially the samples of A and B have equal number of nuclei. After 80 min, the ratio of remaining number of A and B nuclei is:

1 : 16

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4 : 1

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1 : 4

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1 : 1

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Solution

The correct option is C 1 : 4
$\begin{matrix} t_{A} = 20 m i n & t_{B} = 40 m i n n_{A} = \frac{T}{t_{A}} = \frac{80}{20} = 4 & n_{B} = \frac{80}{40} = 2 N_{A} = N {(\frac{1}{2})}^{n_{A}} & n_{B} = N {(\frac{1}{2})}^{n_{B}} N_{A} = N_{0} {(\frac{1}{2})}^{4} & N_{B} = N_{0} {(\frac{1}{2})}^{2} \frac{N_{A}}{N_{B}} = \frac{1}{4} \end{matrix}$

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Half lives of two radioactive substances A and B are respectively 20 min and 40 min. Initially the samples of A and B have equal number of nuclei. After 80 min, the ratio of remaining number of A and B nuclei is:

The correct option is C 1 : 4tA=20 mintB=40 minnA=TtA=8020=4nB=8040=2NA=N(12)nAnB=N(12)nBNA=N0(12)4NB=N0(12)2NANB=14