Physics - Modern Physics Question with Solution | TestHub

$R=R_0{e}^{-\lambda t}$

$\Rightarrow \left(\frac{1}{3}\right)=e^{-\lambda \times 3}=e^{-3\lambda } \dots \left(\mathrm{i}\right)$

Again$\frac{R\text{'}}{R_0}=e^{-\lambda \times 9}=e^{-9\lambda }={\left(e^{-3\lambda }\right)}^3$

=${\left(\frac{1}{3}\right)}^3 [\mathrm{F}\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{ }\mathrm{E}\mathrm{q}.\left(\mathrm{i}\right)]$

$=\frac{1}{27}$

$\Rightarrow$$R’ =$$\frac{R_0}{27}$

Hence, in 9 days activity will become$\left(\frac{1}{27}\right)$of the original value.