Question

Two radioactive materials $X_1$ and $X_2$ contain same number of nuclei. If $6\lambda {\text{sec}}^{-1}$ and $4\lambda {\text{sec}}^{-1}$ are the decay constants of $X_1$ and $X_2$ respectively, then the ratio of number of nuclei undecayed of $X_1$ to that of $X_2$ will be $1/e$ after a time:

• A. $\frac{1}{2\lambda }\text{sec}$
• B. $\frac{1}{5\lambda }\text{sec}$
• C. $\frac{1}{8\lambda }\text{sec}$
• D. $\frac{1}{10\lambda }\text{sec}$

Answer 1

The correct option is A $\frac{1}{2\lambda }\text{sec}$

Using equation $N=N_0{e}^{-\lambda t}$, we have

$\frac{N_{X_1}}{N_{X_2}}=\frac{N_0{e}^{-{\lambda }_{X_1}t}}{N_0{e}^{-{\lambda }_{X_2}t}}$

Substituting the values,

$\frac{1}{e}=\frac{e^{-6\lambda t}}{e^{-4\lambda t}}$

$\Rightarrow e^{-1}=e^{-\lambda t(6-4)}$

Equating the power, we get

$1=\lambda t\left(2\right)$

$\therefore t=\frac{1}{2\lambda }\text{sec}$

Hence, $\left(A\right)$ is the correct answer.

Why this problem? It will help the student to improve the skill of solving the little complex model in decay law.