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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
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There are two radioactive substance A and B . Decay constant of B is two times that of A . Initially both have equal no.of nuclei . After n Lay liver of A rate of disintegration of both are equal . The value of n is

$(a)\;4\qquad(b)\;2\qquad(c)\;1\qquad(d)\;5$

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Answer : (c) 1
Explanation :
Let $\;\lambda_{A}=\lambda \quad \;\lambda_{B}=2 \lambda$
If $\;N_{0}\;$ is total no.of atoms in A and B at $\;t=0\;,$ then initial rate of disintegration of $\;A=\lambda N_{0}\;,$ and initial rate of disintegration of $\;B=2 \lambda N_{0}$
As $\;\lambda_{B}=2 \lambda \alpha$
$T_{B}=\large\frac{1}{2}\;T_{\alpha}$
i.e- Lay life of B is lay the lay life of A
$(-\large\frac{dN}{dt})_{A}=\large\frac{\lambda N_{0}}{2}$
Equivalently after two half lives of B
$(-\large\frac{dN}{dt})_{B}=2 \large\frac{2\lambda N_{0}}{4}=\large\frac{\lambda N_{0}}{2}$
Clearly , $\;(-\large\frac{dN}{dt})_{A}=-(\large\frac{dN}{dt})_{B}$
After $\;n=1\;,$ i.e. one half life of A.
answered Feb 24, 2014 by yamini.v
 

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