In a radioactive disintegration, the ratio of initial number of atoms to the number of atoms present at time $t=\frac{1}{2 \lambda}$ is $[\lambda=$ decay constant $]$

Question

Aicharya · Accepted Answer

According to radioactive disintegration law, $\begin{aligned} & \mathrm{N}=\mathrm{N}_0 \mathrm{e}^{-\lambda \mathrm{t}} \ & \frac{\mathrm{N}}{\mathrm{N}_0}=\mathrm{e}^{-\lambda t} \ & \frac{\mathrm{N}}{\mathrm{N}_0}=\mathrm{e}^{-\lambda 	imes \frac{1}{2 \lambda}} \ & \left(\because t=\frac{1}{2 \lambda}ight) \ & \frac{\mathrm{N}}{\mathrm{N}_0}=\mathrm{e}^{-\frac{1}{2}} \ & \frac{\mathrm{N}_0}{\mathrm{~N}}=\mathrm{e}^{\frac{1}{2}} \ & \frac{\mathrm{N}_0}{\mathrm{~N}}=\sqrt{\mathrm{e}} \ & \end{aligned}$

In a radioactive disintegration, the ratio of initial number of atoms to the number of atoms present at time…

Solution