Question

Two radioactive material $A_1$ and $A_2$ have decay constants of $10{\lambda }_0$ and ${\lambda }_0$. If initially they have same number of nuclei, the ratio of number of their undecayed nuclei will be (1/e) after a time

• A. $\frac{1}{{\lambda }_0}$
• B. $\frac{1}{9{\lambda }_0}$
• C. $\frac{1}{10{\lambda }_0}$
• D. 1

Answer 1

Answer: (B)

$\frac{1}{9{\lambda }_0}$

$\begin{array}{c}\text{A, have decay constant}={10\lambda }_0\\ {\text{A}}_2\text{have decay constent}={\lambda }_0\end{array}$

$\text{Intially they have equal no of nuclei.}$

$\text{we know, after time t,}$

$N_t=N_0{e}^{-\lambda t}$

$\text{for}{A}_1\cdots >N_{1t}=N_0{e}^{-10{\lambda }_{0}t}$

$\text{for}{A}_2\cdots >N_{2t}=N_0{e}^{-{\lambda }_{0}t}$

$\begin{array}{c}\begin{array}{c}\text{taking ratio,}\\ \frac{N_{1t}}{N_{2t}}=e^{-9{\lambda }_{0}t}\\ \text{since ratio is given as}\frac{1}{e}\\ \frac{1}{e}=e^{-9{\lambda }_{0}t}\\ \text{or}=-9{\lambda }_{0}t=-1\Rightarrow t=\frac{1}{9{\lambda }_0}\end{array}\end{array}$