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 un disintegrated nuclei are same is

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

Answer 1

Answer: (C)

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

Answer 2

: $N_{A}e^{- \lambda}A^{t} = N_{B}{e^{- \lambda}}_{B^{t}}$

Or $e^{(\lambda_{B} - \lambda_{A})t} = (N_{B}/N_{A})$

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