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$${ \left( \cfrac { d{ N }_{ 0 } }{ dt } \right) }_{ A }={ \left( \cfrac { d{ N }_{ 0 } }{ dt } \right) }_{ B }$$

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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