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# Which sample, $$A$$ or $$B$$ shown in figure has shorter mean-life?

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#### Initially at $$t=0$$ from figure given$${ \left( \cfrac { d{ N }_{ 0 } }{ dt } \right) }_{ A }={ \left( \cfrac { d{ N }_{ 0 } }{ dt } \right) }_{ B }$$So $${({N}_{0})}_{A}={({N}_{0})}_{B}$$ie initially both samples has equal number of radioactive atoms. Considering at any instant $$t=t$$ from figure.$${ \left( \cfrac { d{ N }_{ 0 } }{ dt } \right) }_{ A }>{ \left( \cfrac { d{ N }_{ 0 } }{ dt } \right) }_{ B }\quad$$$${ \lambda }_{ A }{ N }_{ A }={ \lambda }_{ B }{ N }_{ B }$$$${ N }_{ A }>{ N }_{ B }$$$${ \lambda }_{ A }<{ \lambda }_{ B }\left( \cfrac { -dN }{ dt } =\lambda N \right)$$$$\quad \therefore \tau =\cfrac { 1 }{ \lambda }$$$$\therefore \cfrac { 1 }{ { \tau }_{ A } } <\cfrac { 1 }{ { \tau }_{ B } }$$$${ \tau }_{ A }>{ \tau }_{ B }$$So mean life time of sample $$A$$ is greater than of $$B$$

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