Question

Two radioactive nuclei $A$ and $B$ are taken with their disintegration constant $\lambda_{A}$ and $\lambda_{B}$ and initially $N_{A}$ and $N_{B}$ number of nuclei are taken then the time after which their undisintegrated nuclei are same is

• A. $\frac{\lambda_{A}\lambda_{B}}{\left(\lambda_{A}-\lambda_{B}\right)}$ ln $\left(\frac{N_{B}}{N_{A}}\right)$
• B. $\frac{1}{\left(\lambda_{A}+\lambda_{B}\right)}$ ln $\left(\frac{N_{B}}{N_{A}}\right)$
• C. $\frac{1}{\left(\lambda_{B}-\lambda_{A}\right)}$ ln $\left(\frac{N_{B}}{N_{A}}\right)$
• D. $\frac{1}{\left(\lambda_{A}-\lambda_{B}\right)}$ ln $\left(\frac{N_{B}}{N_{A}}\right)$

Answer 1

Answer: (C)

$\frac{1}{\left(\lambda_{B}-\lambda_{A}\right)}$ ln $\left(\frac{N_{B}}{N_{A}}\right)$

Answer 2

$N_{A} e^{-\lambda_{A^t}}=N_{B^{e^{-\lambda_{B^t}}}}$ or $e^{\left(\lambda_{B}-\lambda_{A}\right)t}=\left(N_{B} N_{A}\right)$ $\therefore t=\frac{1}{\left(\lambda_{B}-\lambda_{A}\right)}$ ln $\left(\frac{N_{B}}{N_{A}}\right)$