Get Answers to all your Questions

header-bg qa

The activity of a radioactive sample falls from 700 \mathrm{s}^{-1} to  500 \mathrm{s}^{-1} in 30 minutes. Its half life ( in minutes ) is close to :
 
Option: 1 62
Option: 2 72
Option: 3 66
Option: 4 52
 

Answers (1)

best_answer

Activity at any time is given as

A=A_0e^{-\lambda t}\\ \Rightarrow \frac{A}{A_0}=e^{-\lambda t}\\ \Rightarrow ln( \frac{A_0}{A})=\lambda t

After half-life,

 \\t=T_{\frac{1}{2}} \\ A=\frac{A_0}{2}

so \ln 2=\lambda T_{1 / 2}

Now for t=30 \ min \ , A_0=700 \ s^{-1}, \ A=500 \ s^{-1}

\ln \left[\frac{700}{500}\right]=\lambda(30 )

from( (i) and (ii)

\frac{\ln 2}{\ln (7 / 5)}=\frac{T_{1 / 2}}{(30 )}

{t_{1 / 2}} =61.81 \cong 62 \ min

Hence the correct option is (1). 

Posted by

Ritika Jonwal

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE