Question

Two radioactive materials X ${}_1$and X${}_2$ have decay constants 10$\lambda$ and $\lambda$ respectively. If finally they have the same number of nuclei, then the ratio of the number of nuclei of X ${}_1$ to that of X${}_2$ will be 1/e after a time

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

Answer 1

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

Note:- time is not given when both have same no of nuclei
Given that, finally after time t, both are left with same no of nuclei.

Initial in $X_1=N_1$

Initial in $x_2=N_2$

$N_1{e}^{-10\lambda t}=N_2{e}^{-\lambda t}$

$ln\left(\frac{N_1}{N_2}\right)-10\lambda t=-\lambda t$

$\Rightarrow 9\lambda t=ln\left(\frac{N_1}{N_2}\right)$

$\Rightarrow t=\frac{ln\left(\frac{N_1}{N_2}\right)}{9\lambda }$

$\frac{1}{e}=\frac{N_1{e}^{-10\lambda t}}{N_2{e}^{-\lambda t}}\Rightarrow \frac{N_2}{N_1}{e}^9\lambda t=e$

Taking $ln$, $ln\left(\frac{N_2}{N_1}\right)+9\lambda t=1$