Question

A radioactive nucleus $A$ with a half-life $T$ , decays into a nucleus $B$ . At $t=0,$ there is no nucleus $B$ . At some time $t$ , the ratio of the number of $B$ to that of $A$ is $0.3.$ Then, $t$ is given by

• A. $\frac{T}{\log 1 . 3}$
• B. $\frac{T}{2}\frac{\log 2}{\log 1 . 3}$
• C. $T\frac{\log 1 . 3}{\log 2}$
• D. $T\log1.3$

Answer 1

Answer: (C)

$A \rightarrow B$
At $t=0,$ Number of nuclei of $A$ be $N_{0}$
At $t=t,$ Number of nuclei of $A$ remaining, $N_{A}=N_{0}e^{- \lambda t}$
Number of nuclei of $B$ formed, $N_{B}=N_{0}\left(1 - e^{- \lambda t}\right)$
$\frac{N_{B}}{N_{A}}=\frac{1 - e^{- \lambda t}}{e^{- \lambda t}}=0.3$
$\Rightarrow 1-e^{- \lambda t}=0.3e^{- \lambda t}$
$\Rightarrow 1=1.3e^{- \lambda t}$
$\Rightarrow \left(1 . 3\right)^{- 1}=e^{- \lambda t}$
$\Rightarrow ln\left(1 . 3\right)^{- 1}=ln\left(e^{- \lambda t}\right)$
$\Rightarrow ln\left(1 . 3\right)=\lambda t$
$\Rightarrow ln\left(1 . 3\right)=\frac{ln \left(2\right)}{T}t$
$\Rightarrow t=T\frac{\log \left(1 . 3\right)}{\log \left(2\right)}$