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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

Last updated date: 24th Jun 2024
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Answer
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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
$\text{t1/2}=\dfrac{\text{ln 2}}{\text{ }\!\!\lambda\!\!\text{ }}$