Question

Two radioactive material $X_1$ and $X_2$ have decay constant $10\lambda$ and $\lambda$ respectively. Initially, they have the same number of nuclei. The ratio of the number of nuclei $X_1$ to that of $X_2$ will be $\frac{1}{e}$ after time,

• A. $\frac{1}{10\lambda }$
  
• B. $\frac{1}{11\lambda }$
  
• C. $\frac{1}{100\lambda }$
  
• D. $\frac{1}{9\lambda }$
  

Answer 1

The correct option is D $\frac{1}{9\lambda }$


For $X_1$, $N_1=N_0{e}^{-{\lambda }_{1}t}$
For $X_2$, $N_2=N_0{e}^{-{\lambda }_{2}t}$
Here, ${\lambda }_1=10\lambda$, ${\lambda }_2=\lambda$

$\therefore \frac{N_1}{N_2}=\frac{N_0{e}^{-{\lambda }_{1}t}}{N_0{e}^{-{\lambda }_{2}t}}$

$\frac{1}{e}=\frac{e^{-{\lambda }_{1}t}}{e^{-{\lambda }_{2}t}}$

$e^{-1}=e^{-({\lambda }_1-{\lambda }_2)t}$

$1=({\lambda }_1-{\lambda }_2)t$

$t=\frac{1}{{\lambda }_1-{\lambda }_2}=\frac{1}{10\lambda -\lambda }=\frac{1}{9\lambda }$

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