Physics - Modern Physics Question with Solution | TestHub

Let active Nuclei at $t=0$ is $N_0$ then, active nuclei at $t=10$ hour $N_1=N_0{e}^{-10\lambda }$ Active nuclei at $t=11$ hour $N_2=N_0{e}^{-11\lambda }$ $\begin{array}{rl}& \%\text{ decay }=\left(\frac{N_1-N_2}{N_1}\right)\times 100\\ & =\left(\frac{N_0\cdot e^{-10\lambda }-N_0{e}^{-11\lambda }}{N_1}\right)\times 100\\ & =\left(\frac{N_0\cdot e^{-10\lambda }-N_0{e}^{-11\lambda }}{N_0\cdot e^{-10\lambda }}\right)\times 100\\ & =\left(1-\frac{1}{e^{\lambda }}\right)\times 100=\left(1-\frac{1}{e^{\ln 2/T_{1/2}}}\right)\times 100\\ & =\left(\because T_{1/2}=69.3\mathrm{hrs}.\right)\\ & =\left(1-\frac{1}{e^{0.01}}\right)\times 100\Rightarrow \%\text{ decay }=1\%.\end{array}$