Question

Two radioactive substances $A$ and $B$ have decay constants $5\lambda$ and $\lambda$ respectively. At $t=0$, they have the same number of nuclei. The ratio of number of nuclei of $A$ to those of $B$ will be $(1/e{)}^2$ after a time interval

• A. $4\lambda$
• B. $2\lambda$
• C. $1/2\lambda$
• D. $1/4\lambda$

Answer 1

Answer: (C)

Given : ${\lambda }_A=5\lambda ,{\lambda }_B=\lambda$
At $t=0,{\left(N_0\right)}_A={\left(N_0\right)}_B$
At time $t,\frac{N_A}{N_B}={\left(\frac{1}{e}\right)}^2$
According to radioactive decay, $\frac{N}{N_0}=e^{-\lambda t}$
$\therefore \frac{N_A}{{\left(N_0\right)}_A}=e^{-{\lambda }_{A^t}}$
and $\frac{N_B}{{\left(N_0\right)}_B}=e^{-{\lambda }_{B^t}}$
Divide $\left(i\right)$ by $\left(ii\right)$, we get
$\frac{N_A}{N_B}=e^{-\left({\lambda }_A-{\lambda }_B\right)t}$ or $\frac{N_A}{N_B}=e^{-\left(5\lambda -\lambda \right)t}$
or ${\left(\frac{1}{e}\right)}^2=e^{-4\lambda t}$ or ${\left(\frac{1}{e}\right)}^2={\left(\frac{1}{e}\right)}^{4\lambda t}$
$\Rightarrow 4\lambda t=2$ or $t=\frac{2}{4\lambda }=\frac{1}{2\lambda }$