The activity of a radioactive material is 2.56 $\times$ 10^$-$3 Ci. If the half life of the material is 5 days, after how many days the activity will become 2 $\times$ 10^$-$5 Ci ?

By Radioactive Decay law, <br/><br/>$$ \begin{aligned} & \mathrm{R}=\mathrm{R}_{\mathrm{o}} e^{-\lambda t} \\\\ & \Rightarrow 2 \times 10^{-5}=2.56 \times 10^{-3} e^{-\lambda t} \end{aligned} $$ <br/><br/>[Where, $\mathrm{R} =$ Activity at time $t$ <br/><br/>$\lambda =$ Activity constant of Radioactive sample <br/><br/>and $\lambda = \frac{\ln 2}{\mathrm{~T}_{1 / 2}}$ <br/><br/>Taking logarithm on both sides <br/><br/>$$\ln \left(2 \times 10^{-5}\right)=\ln \left(2.56 \times 10^{-3}\right)+\ln \left(e^{-\lambda t}\right)$$ <br/><br/>$$\Rightarrow\ln \left(2 \times 10^{-5}\right)-\ln \left(2.56 \times 10^{-3}\right)=-\lambda t$$ <br/><br/>$\Rightarrow \ln \left(\frac{2 \times 10^{-5}}{2.56 \times 10^{-3}}\right)=-\lambda t$ <br/><br/>$\Rightarrow \ln \left(\frac{1}{128}\right)=-\lambda t$ <br/><br/>$\Rightarrow-\ln 128=-\lambda t$ <br/><br/>$\Rightarrow \ln 2^7=\frac{\ln 2}{\mathrm{~T}_{1 / 2}} t$ <br/><br/>$\Rightarrow 7 \ln 2=\frac{\ln 2}{\mathrm{~T}_{1 / 2}} t$ <br/><br/>$\Rightarrow t=7 \mathrm{~T}_{1 / 2}$ <br/><br/>$\Rightarrow t=7 \times 5=35$ days