The half-life of a radioactive substance is 20 minutes. The approximate time interval $ \left( {{\text{t}}_{2}}-{{\text{t}}_{1}} \right) $ between time $ {{\text{t}}_{1}} $ when $ \dfrac{1}{3} $ of it had decayed and $ {{\text{t}}_{2}} $ when $ \dfrac{2}{3} $ of it had decayed is:
(A) 7 min
(B) 14 min
(C) 20 min
(D) 28 min

Hint: The radioactivity formula is given by:
 $ \text{N}={{\text{N}}_{0}}\text{ }{{\text{e}}^{-\text{ }\!\!\lambda\!\!\text{ t}}} $
Where $ {{\text{N}}_{0}} $ = initial substance at t = 0
N = substance left at time t
By substituting the given conditions in this equation, the answer can be calculated.

Complete step by step solution:
We know that the formula for radioactive decay is given by:
 $ \text{N}={{\text{N}}_{0}}\text{ }{{\text{e}}^{-\text{ }\!\!\lambda\!\!\text{ t}}} $ …… (1)
Now, in time $ {{\text{t}}_{1,}}\dfrac{1}{3} $ rd of the radioactive substance has been designed. So substance left is $ 1-\dfrac{1}{3}=\dfrac{2}{3}\text{rd} $ .
So, $ \text{N}=\dfrac{2}{3}{{\text{N}}_{0}} $
Therefore, putting this value in equation (1), we get,
 $ \begin{align}
  & =\dfrac{2}{3}{{\text{N}}_{0}}={{\text{N}}_{0}}\text{ }{{\text{e}}^{-\text{ }\!\!\lambda\!\!\text{ t}}} \\
 & \dfrac{2}{3}\text{= }{{\text{e}}^{-\text{ }\!\!\lambda\!\!\text{ t}}} \\
 & \text{ln}\dfrac{2}{3}=-\text{ }\!\!\lambda\!\!\text{ }{{\text{t}}_{1}} \\
 & 2\cdot 303\text{ log}\dfrac{2}{3}=-\text{ }\!\!\lambda\!\!\text{ }{{\text{t}}_{1}} \\
\end{align} $
 $ \begin{align}
  & {{\text{t}}_{1}}=\dfrac{-2\cdot 303\text{ log}\left( \dfrac{2}{3} \right)}{\text{ }\!\!\lambda\!\!\text{ }} \\
 & {{\text{t}}_{1}}=\dfrac{0\cdot 40553817}{\text{ }\!\!\lambda\!\!\text{ }} \\
\end{align} $
In time $ {{\text{t}}_{2,}}\dfrac{2}{3}\text{rd} $ of the radioactive substance has decayed. So, substance left is $ 1-\dfrac{2}{3}=\dfrac{1}{3}\text{rd} $
So, $ \text{N}=\dfrac{1}{3}{{\text{N}}_{0}} $
Therefore, putting this value in equation (1), we get
 $ \dfrac{1}{3}{{\text{N}}_{0}}={{\text{N}}_{0}}{{\text{e}}^{-\text{ }\!\!\lambda\!\!\text{ }{{\text{t}}_{2}}}} $
 $ \dfrac{1}{3}={{\text{e}}^{-\text{ }\!\!\lambda\!\!\text{ }{{\text{t}}_{2}}}} $
 $ \text{ln}\dfrac{1}{3}-\text{ }\!\!\lambda\!\!\text{ }{{\text{t}}_{\text{2}}} $
 $ 2\cdot 303\text{ log}\dfrac{1}{3}=-\text{ }\!\!\lambda\!\!\text{ }{{\text{t}}_{2}} $
 $ \begin{align}
  & {{\text{t}}_{2}}=\dfrac{-2\cdot 303\text{ log}\left( \dfrac{1}{3} \right)}{\text{ }\!\!\lambda\!\!\text{ }} \\
 & {{\text{t}}_{2}}=\dfrac{1\cdot 09881025}{\text{ }\!\!\lambda\!\!\text{ }} \\
\end{align} $
Now, $ {{\text{t}}_{2}}-{{\text{t}}_{1}}=\dfrac{\left( 1\cdot 09881025-0\cdot 40553817 \right)}{\text{ }\!\!\lambda\!\!\text{ }} $
 $ {{\text{t}}_{2}}-{{\text{t}}_{1}}=\dfrac{0\cdot 69327208}{\text{ }\!\!\lambda\!\!\text{ }} $ …. (2)
Now we know that half-life is given by
 $ {{\text{t}}_{2}}-{{\text{t}}_{1}}=\dfrac{0\cdot 693}{\text{ }\!\!\lambda\!\!\text{ }} $
Or $ \text{ }\!\!\lambda\!\!\text{ }=\dfrac{0\cdot 693}{\text{t1/2}} $
as $ \text{t1/2} $ =20 minutes
So $ \text{ }\!\!\lambda\!\!\text{ }=\dfrac{0\cdot 693}{20} $ ….. (3)
Putting value of (3) in (2), we get
 $ \begin{align}
  & {{\text{t}}_{2}}-{{\text{t}}_{1}}=\dfrac{0\cdot 69327208}{0\cdot 693}\times 20 \\
 & \text{ }=20\cdot 0078522 \\
 & {{\text{t}}_{2}}-{{\text{t}}_{1}}=20\text{ minutes} \\
\end{align} $
So , the correct option is (C) .

Note:
Radioactive decay is the breakdown of atomic nucleus which results in release of energy and matter from the nucleus.
The law of radioactive decay describes the statistical behaviour of a large number of nuclides, rather than individual ones.
Given a sample, the number of decay events $ -\text{dN} $ in small interval dt is proportional to number of atoms N present, that is:
 $ \dfrac{-\text{dN}}{\text{dt}}\alpha \text{ N} $
Or $ \dfrac{-\text{dN}}{\text{N}}=\text{ }\!\!\lambda\!\!\text{ dt} $
Where $ \text{ }\!\!\lambda\!\!\text{ } $ =decay constant. On integrating, we get
 $ \text{N}={{\text{N}}_{0}}\text{ }{{\text{e}}^{-\text{ }\!\!\lambda\!\!\text{ t}}} $
This equation is called the decay rate equation. The half-life is related to decay constant by the formula
 $ \text{t1/2}=\dfrac{\text{ln 2}}{\text{ }\!\!\lambda\!\!\text{ }} $