First slide

Radio activity

Question

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

Moderate

A

1 : 16
B

4 : 1
C

1 : 4
D

1 : 1

Solution

For 80 minutes, number of half-lives of sample
$A = n_{A} = \frac{80}{20} = 4$ and number of half-lives of sample
$B = n_{B} = \frac{80}{40} = 2 .$ Also by using $N = N_{0} {(\frac{1}{2})}^{n}$
$\Rightarrow N \propto \frac{1}{2^{n}} \Rightarrow \frac{N_{A}}{N_{B}} = \frac{2^{nB}}{2^{nA}} = \frac{2^{2}}{2^{4}} = \frac{1}{4}$

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