For a radioactive material, its activity A and rate of change of...

N = {N}_{0}{e}^{- \lambdat}$<br >$A = \lambda{N}_{0}{e}^{- \lambdat}$<br >$R = {\lambda}^{2}{N}_{0}{e}^{- \lambdat}$<br >${A}_{P} = {A}_{Q}$at$t = 0$<br >${\lambda}_{P}{N}_{P}{e}^{- {\lambda}_{P}t} = {\lambda}_{Q}{N}_{Q}{e}^{- {\lambda}_{Q}t}$at$t = 0$<br >${\lambda}_{P}{N}_{P} = {\lambda}_{Q}{N}_{Q}$...(i) <br >$\dfrac{{R}_{P}}{{R}_{Q}} = {\left(\dfrac{{\lambda}_{P}}{{\lambda}_{Q}}\right)}^{2}$\left(\dfrac{{N}_{P}}{{N}_{Q}}\right)\dfrac{{e}^{- {\lambda}_{P}2τ}}{{e}^{- {\lambda}_{Q}2τ}}$
$\dfrac{{R}_{P}}{{R}_{Q}} = {\left(\dfrac{{\lambda}_{P}}{{\lambda}_{Q}}\right)}^{2}\left(\dfrac{{N}_{P}}{{N}_{Q}}\right)\dfrac{1}{e}$ ...(ii)
\left[{\lambda}_{P} = \dfrac{1}{meanlife}\right]$<br >From equation (i) and equation (ii)<br >$\dfrac{{R}_{P}}{{R}_{Q}} = \dfrac{2}{e} ⇒ n = 2