Question

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

• A. $t=\frac{T}{\log \left(1.3\right)}$
• B. $t=\frac{T}{2}\frac{\log 2}{\log 1.3}$
• C. $t=T\frac{\log 1.3}{\log 2}$
• D. $t=T\log \left(1.3\right)$

Answer 1

The correct option is C $t=T\frac{\log 1.3}{\log 2}$

Since the radioactive nucleus has half life $T$,
$\Rightarrow \lambda =\frac{ln2}{T}$
$A\underset{\lambda }{\overset{}{\to }}B$

At $t=0$, Number of $A=N_0$ and Number of $B=0$
Therefore at time $=t$,

Number of $A=N_0{e}^{-\lambda t}$

Number of $B=N_0-N_0{e}^{-\lambda t}=N_0(1-e^{-\lambda t})$

$\Rightarrow \frac{\text{Number of}B}{\text{Number of}A}=\frac{N_0(1-e^{-\lambda t})}{N_0{e}^{-\lambda t}}=0.3$

$\Rightarrow e^{\lambda t}-1=0.3$

$\Rightarrow e^{\lambda t}=1.3\Rightarrow \lambda t=ln\left(1.3\right)$

$\Rightarrow t=\frac{ln1.3}{\lambda }=T\frac{ln1.3}{ln2}$
$=\frac{Tlog1.3}{log2}$