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

Answer: (C)

$t=T\frac{\log 1.3}{\log 2}$

Since the radioactive nucleus has half life ${}^{\prime }{T}^{\prime }$,

$\Rightarrow \lambda =\frac{ln2}{T}$

$A\underset{\lambda }{\overset{}{\to }}B$

At $t=0$, No. of $A=N_2$ and No. of $B=0$

Therefore at time $=t$,

$No.ofA=N_0{e}^{-\lambda t}$

$No.ofB=N_0-N_0{e}^{-\lambda t}=N_0(1-e^{-\lambda t})$

$\Rightarrow \frac{No.ofB}{No.ofA}=\frac{N_0(1-e^{-\lambda t})}{N_0{e}^{\lambda t}}=0.3$

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

$\Rightarrow e^{\lambda t}=1.3\Rightarrow \lambda t=ln1.3$

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

$=\frac{Tlog1.3}{log2}$

Hence Correct answer is option (C).