Question 463: Atoms & Nuclei

Question

Two radioactive materials $$X_1$$ and $$X_2$$ have decay constants '$$5 \lambda$$' and '$$\lambda$$' respectively. Initially, they have the same number of nuclei. After time '$$t$$', the ratio of number of nuclei of $$X_1$$ to that of $$\mathrm{X}_2$$ is $$\frac{1}{\mathrm{e}}$$. Then $$\mathrm{t}$$ is equal to

Accepted Answer

To solve for time $$t$$ based on the given conditions, let's apply the law of radioactive decay. The number of nuclei $$N$$ remaining after time $$t$$ from an initial amount $$N_0$$ for a substance with a decay constant $$\lambda$$ is described by the equation:
$$N = N_0 e^{-\lambda t}$$
Given two radioactive materials $$X_1$$ and $$X_2$$ with decay constants $$5\lambda$$ and $$\lambda$$ respectively, and the same initial number of nuclei, $$N_0$$, the number of remaining nuclei at time $$t$$ for each can be described as follows:
For $$X_1$$:
$$N_1 = N_0 e^{-5\lambda t}$$
For $$X_2$$:
$$N_2 = N_0 e^{-\lambda t}$$
The given ratio of the number of nuclei of $$X_1$$ to $$X_2$$ after time $$t$$ is $$\frac{1}{e}$$. Therefore, we can set up the equation:
$$\frac{N_1}{N_2} = \frac{N_0 e^{-5\lambda t}}{N_0 e^{-\lambda t}} = \frac{1}{e}$$
Simplifying the expression, we cancel out $$N_0$$:
$$e^{\lambda t} = e^{5\lambda t} \cdot \frac{1}{e}$$
taking natural logarithm on both side
$$\implies \lambda t = 5 \lambda t - 1$$
Upon rearranging the equation, to solve for $$t$$, we find:
$$-4\lambda t = -1$$
Solving for $$t$$:
$$t = \frac{1}{4\lambda}$$
Thus, the correct option is D: $$\frac{1}{4\lambda}$$.

Answer

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

Answer

$$\frac{\mathrm{e}}{\lambda}$$

Answer

$$\lambda$$

Answer

$$\frac{1}{4 \lambda}$$

Atoms & Nuclei

Two radioactive materials $X_1 and X_2 have decay constants '5 \lambda' and '\lambda' respectively. Initially, they have the same number of nuclei. After time 't', the ratio of number of nuclei of X_1 to that of \mathrm{X}_2 is \frac{1}{\mathrm{e}}. Then \mathrm{t}$ is equal to

Choose the correct answer: