In a radioactive decay process, the activity is defined as A = -...

Activity for both the cases can be written as, ${A}_{1} = {A}_{0}{e}^{- {\lambda}_{1}{t}_{1}}$

also ${A}_{2} = {A}_{0}{e}^{- {\lambda}_{2}{t}_{2}}$.

Half-life for the reaction can be written as, ${\left(t\dfrac{1}{2}\right)}_{1} = \dfrac{\ln 2}{{\lambda}_{1}}$ and hence

at ${t}_{1} = \dfrac{3\ln 2}{{\lambda}_{1}}$.

${A}_{1} = {A}_{0}{e}^{- {\lambda}_{1}\left(\dfrac{3\ln 2}{{\lambda}_{1}}\right)}$

$⇒ {A}_{1} = {A}_{0}{e}^{- 3\ln 2} . . . \left(i\right)$

Similarly, for the second case, we can write

${t}_{2} = 7 \left(\dfrac{\ln 2}{{\lambda}_{2}}\right) and therefore,</p><p>${A}_{2} = {A}_{0}{e}^{- {\lambda}_{2}\left(\dfrac{7\ln 2}{{\lambda}_{2}}\right)}

${A}_{2} = {A}_{0}{e}^{- 7\ln 2} . . . \left(ii\right)$

From \left(i\right)$and \left(ii\right)$, we get

$\dfrac{{A}_{1}}{{A}_{2}} = \dfrac{{A}_{0}{e}^{- 3\ln 2}}{{A}_{0}{e}^{- 7\ln 2}} = \dfrac{{A}_{0}{e}^{ln{2}^{- 3}}}{{A}_{0}{e}^{ln{2}^{- 7}}} = \dfrac{{2}^{- 3}}{{2}^{- 7}} = 16$

Therefore, required ration would be$∴ \dfrac{{A}_{1}}{{A}_{2}} = 16$