### JEE Main On 16 April 2018 Question 5

##### Question: At some instant, a radioactive sample $ S _1 $ having an activity $ 5\mu Ci $ has twice the number of nuclei as another sample $ S _2 $ which has an activity of $ 10\mu Ci. $ The half-lives $ S _1 $ of and $ S _2 $ are [JEE Main 16-4-2018]

#### Options:

A) 10 years and 20 years, respectively

B) 5 years and 20 years, respectively

C) 20 years and 10 years, respectively

D) 20 years and 5 years, respectively

## Show Answer

#### Answer:

Correct Answer: B

#### Solution:

For sample 1 we can write,

$ N _1=N _{o1}{e^{-{\lambda_1}t}} $ $ N _2=N _{o2}{e^{-{\lambda_2}t}} $

Given that at a particular time t $ N _1=2N _2 $

Also given that For sample 1, $ A _1=-\frac{dN _1}{dt}=N _{o1}{\lambda_1}{e^{-{\lambda_1}t}}=5\mu C _{i} $

$ A _2=-\frac{dN _2}{dt}=N _{o2}{\lambda_2}{e^{-{\lambda_2}t}}=10\mu C _{i} $

Solving above equations we get, $ \frac{{\lambda_1}}{{\lambda_2}}=\frac{1}{4} $ $ \frac{half,life,e _1}{half,life,e _2}=\frac{\ln 2}{{\lambda_1}}\times \frac{{\lambda_2}}{\ln 2}=4 $

It is clear that the ratio of half lives of the two samples is

4:1, therefore half lives can be 20(4k) years and 5(1k) years respectively.