Question 442: Atoms & Nuclei

Question

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]

Accepted Answer

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

Answer

$$\frac{1}{\mathrm{e}}$$

Answer

$$\sqrt{\mathrm{e}}$$

Answer

e

Answer

2e

Atoms & Nuclei

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]

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