Question 64: Atoms & Nuclei

Question

Two radioactive materials X$$_1$$ and X$$_2$$ have decay constant 5$$\lambda$$ and $$\lambda$$, respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X$$_1$$ to X$$_2$$ will be $$1/e$$ after a time

Accepted Answer

Decays constants of radioactive materials $$X_1$$ and $$X_2$$ is given as $$\lambda_1=5\lambda$$ and $$\lambda_2=\lambda$$
If $$N_0$$ be the initial number of nuclei in both radioactive materials, then according to radioactive decay law,
$${N_{{X_1}}} = {N_0}{e^{ - {\lambda _1}t}}$$
and $${N_{{X_2}}} = {N_0}{e^{ - {\lambda _2}t}}$$
$$	herefore$$ $${{{N_{{X_1}}}} \over {{N_{{X_2}}}}} = {{{N_0}{e^{ - {\lambda _1}t}}} \over {{N_0}{e^{ - {\lambda _2}t}}}}$$
$$\frac{1}{e}=e^{\left(-\lambda_1+\lambda_2ight) t}$$
$$\begin{aligned} & {\left[	ext { Given, } \frac{N_{X_1}}{N_{X_2}}=\frac{1}{e}ight]} \ & \Rightarrow \quad e^{-1}=e^{-\left(\lambda_1-\lambda_2ight) t} \ & \Rightarrow \quad\left(\lambda_1-\lambda_2ight) t=1 \ & \Rightarrow \quad t=\frac{1}{\lambda_1-\lambda_2} \ & =\frac{1}{5 \lambda-\lambda}=\frac{1}{4 \lambda} \ & \end{aligned}$$

Answer

$$1/4\lambda$$

Answer

$$e/\lambda$$

Answer

$$\lambda$$

Answer

$$\frac{1}{2}\lambda$$

Atoms & Nuclei

Two radioactive materials X$_1 and X_2 have decay constant 5\lambda and \lambda, respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X_1 to X_2 will be 1/e$ after a time

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