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) $t = \dfrac{T}{{\log (1.3)}}$
B) $t = \dfrac{T}{2}\dfrac{{\log 2}}{{\log (1.3)}}$
C) $t = T\dfrac{{\log (1.3)}}{{\log 2}}$
D) $t = T\log (1.3)$

Answer 1

Hint: Radioactive decay happens exponentially. An amount of a radioactive element decays to another element in some interval. When the mother nucleus decays its half amount of nucleus, that time is called its half-life. Here each nucleus of $A$ decays to a nucleus of $B$ so keeping track of the number of the elements is dependent on each other linearly.

Formulae Used:
If a radioactive element has initially ${N_0}$ number of nucleus and after decaying for $t$ time the amount of the element left ${N_t}$ then
${N_t} = {N_0}{e^{ - \lambda t}}$
where, $\lambda $ is the decay constant of that element

Complete Step by Step Answer:
Given:
The radioactive nucleus $A$ has a half-life $T$.
 Each radioactive nucleus $A$ decays into a nucleus $B$.
At $t = 0$ , there is no nucleus $B$.
At the time $t$, the ratio of the number of $B$ to that of $A$ is $0.3$.
To get: The value of $t$.
Step 1:
At time $t = 0$ there was no nucleus of $B$.
Hence, initially there was ${N_0}$ nucleus of $A$.
Now calculate the value of the half-life $T$ of the radioactive element $A$ from eq (1)
$
  \dfrac{{{N_0}}}{2} = {N_0}{e^{ - \lambda T}} \\
   \Rightarrow \dfrac{1}{2} = {e^{ - \lambda T}} \\
   \Rightarrow \log \left( {\dfrac{1}{2}} \right) = \log \left( {{e^{ - \lambda T}}} \right) \\
   \Rightarrow - \log 2 = - \lambda T \\
 $
$\Rightarrow T = \dfrac{{\log 2}}{\lambda }$
Similarly you get the decay constant to be $\lambda = \dfrac{{\log 2}}{T}$
Step 2:
Now, from eq (1) at time $t$ the number of nucleus of $A$ left becomes
\[
  {N_t} = {N_0}{e^{ - \lambda t}} \\
  {N_t} = {N_0}{e^{ - \dfrac{{\log 2}}{T}t}} \\
 \]
Now the current number of $B$ nucleus is the same as the amount of nucleus decayed in the process.
Number of nucleus of $B$ at time $t$ is $\left( {{N_0} - {N_t}} \right) = \left( {{N_0} - {N_0}{e^{ - \dfrac{{\log 2}}{T}t}}} \right)$
Step 3:
By the problem at time $t$, the ratio of the number of $B$ to that of $A$ is $0.3$.
$
\Rightarrow \dfrac{{\left( {{N_0} - {N_0}{e^{ - \dfrac{{\log 2}}{T}t}}} \right)}}{{{N_0}{e^{ - \dfrac{{\log 2}}{T}t}}}} = 0.3 \\
   \Rightarrow {e^{\dfrac{{\log 2}}{T}t}} - 1 = 0.3 \\
   \Rightarrow {e^{\dfrac{{\log 2}}{T}t}} = 1.3 \\
   \Rightarrow \log \left( {{e^{\dfrac{{\log 2}}{T}t}}} \right) = \log \left( {1.3} \right) \\
   \Rightarrow \left( {\dfrac{{\log 2}}{T}t} \right) = \log (1.3) \\
   \Rightarrow t = \dfrac{{T\log \left( {1.3} \right)}}{{\log 2}} \\
 $
$\therefore$ If a radioactive nucleus $A$ with a half life $T$ , decays into a nucleus $B$ and at $t = 0$, there is no nucleus $B$ and again at some time $t$, the ratio of the number of $B$ to that of $A$ is $0.3$, then, $t$ is given by option (C) $t = T\dfrac{{\log (1.3)}}{{\log 2}}$.

Note:
Each nucleus of $A$ decays to a nucleus $B$. There was initially no nucleus of $B$. Hence the number of nuclei $B$ should be only those who have decayed already. The $\log $ operated on both sides are in the base of $e$.