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: