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. $t=(T/2)(\log 2/\log 1.3)$
• B. $t=\left(T\right)(\log 1.3/\log 2)$
• C. $t=T\log (1.3)$
• D. $t=T/[\log (1.3\left)\right]$

Answer 1

The correct option is B

$t=\left(T\right)(\log 1.3/\log 2)$

Step 1: Given data

Half life $=T$

At $t=0$, there is no nucleus $B$

Step 2: Find $t$

Let $\lambda$ be decay constant.

Since the radioactive nucleus has half life $T$ , the decay constant become

$\Rightarrow \lambda =\frac{\ln 2}{T}$

​At $t=0$, there is no nucleus $B$ . So,

Let initial number of atomic nuclei $=N_0$

The number of $A$$=N_0$

The number of $B$$=0$

At some time $t$

The number of $A$$=\left[N_0{e}^{-\lambda t}\right]$

The number of $B$$=[N_0–N_0{e}^{-\lambda t}]$

The ratio of the number of $B$ to that of $A$ is

$$\begin{gathered} [N_0–N_0{e}^{-\lambda t}]/\left[N_0{e}^{-\lambda t}\right]=0.3 \\ {e}^{\lambda t}-1=0.3 \\ {e}^{\lambda t}=1.3 \\ \lambda t=\ln 1.3 \end{gathered}$$

Put value of $\lambda$ , we get,

$$\begin{gathered} (\ln 2/T)t=\ln 1.3 \\ t=T\times (\ln (1.3\left)\right)/\ln 2 \\ t=T\times (\log (1.3\left)\right)/\log 2 \end{gathered}$$

Hence, option B is correct.