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 $\mathrm{0.3.}$ 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. $T\log 1.3$

Answer 1

Answer: (C)

$A\to B$
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, $N_B=N_0\left(1-e^{-\lambda t}\right)$
$\frac{N_B}{N_A}=\frac{1-e^{-\lambda t}}{e^{-\lambda t}}=0.3$
$\Rightarrow 1-e^{-\lambda t}=0.3e^{-\lambda t}$
$\Rightarrow 1=1.3e^{-\lambda t}$
$\Rightarrow {\left(1.3\right)}^{-1}=e^{-\lambda t}$
$\Rightarrow ln{\left(1.3\right)}^{-1}=ln\left(e^{-\lambda t}\right)$
$\Rightarrow ln\left(1.3\right)=\lambda t$
$\Rightarrow ln\left(1.3\right)=\frac{ln\left(2\right)}{T}t$
$\Rightarrow t=T\frac{\log \left(1.3\right)}{\log \left(2\right)}$