Question

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

Answer 1

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 $$B$$