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

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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.

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