Physics - Modern Physics Question with Solution | TestHub

We know that the activity of radioactive substance, $A=A_0{e}^{-\lambda t}$ $$\begin{array}{rl}& \Rightarrow \frac{A}{A_0}=e^{-\lambda t}=e^{-\frac{0.693}{T^{1/2}t}}\left[\because T_{1/2}=\frac{0.693}{\lambda }\right]\\ & \Rightarrow \frac{1}{16}=e^{-\frac{0.693\times 40}{T^{1/2}}}\Rightarrow \frac{0.693\times 40}{T_{1/2}}=\ln 16\\ & \Rightarrow T_{1/2}=\frac{0.693\times 40}{\ln 16}=9.997\approx 10\text{ days. }\end{array}$$