
What is the half-life of a radioactive substance of any given amount of the substance 1M disintegrate in 40 minutes?
A.10 min
B.160 min
C.13 min 20 sec
D.20 min
Answer
514.8k+ views
Hint: Radioactive decay shows disappearance of a constant fraction of activity per unit time. Half-life is the time required to decay a sample to \[50\% \] of its initial activity. In the question,
\[87.5\% \] of any given amount of the substance 1M disintegrate in 40 minutes and thus the remaining amount of the substance is \[12.5\% \]. Substituting the values in the equation, $t = \dfrac{1}{K}In\dfrac{{[{R^\circ }]}}{{[R]}}$, we will calculate the half-life.
Complete step by step answer:
It is given in the question that the \[87.5\% \] of any given amount of the substance 1M disintegrate in 40 minutes.
Let the half-life of the substance be taken to be \[{t_{\dfrac{1}{2}}}\].
$t = \dfrac{1}{K}In\dfrac{{[{R^\circ }]}}{{[R]}}$
Where, \[R^\circ \] is the initial concentration = 1 M
R is the remaining concentration \[ = 12.5\% \]
Substituting the values in the above equation, we get
\[40 = \dfrac{1}{K}In\dfrac{{[{R^\circ }]}}{{\dfrac{{12.5}}{{100}}[R]}}\]
$\Rightarrow$ $K = \dfrac{1}{{40}}In\left( {\dfrac{{100}}{{12.5}}} \right)$
$K = \dfrac{1}{{40}}In\,8$
\[{t_{\dfrac{1}{2}}} = ln\dfrac{2}{K}\]
$ = \dfrac{{In\,2}}{{In\dfrac{5}{{40}}}}$
$ = \dfrac{{In\,2}}{{In\,8}} \times 40$
\[ = \dfrac{{40}}{3}\] minutes
= 13.33 minutes
\[{t_{\dfrac{1}{2}}} = 13\] minutes 20 sec
Hence, the half-life of the substance is 13 min 20 sec.
Therefore, the correct answer is option (C).
Note: The time in which half of the original number of nuclei decay is called as the half-life that is “t sec” in this case. Half of the remaining nuclei decay in the following half-life. Therefore, the number of radioactive nuclei decreases from N to \[\dfrac{N}{2}\] in one half-life and $\dfrac{N}{4}$ in the next and to \[\dfrac{N}{8}\] in the next and so on. Radioactive decay is a spontaneous process. It cannot be predicted exactly for any single nucleus and can only be described statistically and probabilistically i.e. can only give averages and probabilities.
\[87.5\% \] of any given amount of the substance 1M disintegrate in 40 minutes and thus the remaining amount of the substance is \[12.5\% \]. Substituting the values in the equation, $t = \dfrac{1}{K}In\dfrac{{[{R^\circ }]}}{{[R]}}$, we will calculate the half-life.
Complete step by step answer:
It is given in the question that the \[87.5\% \] of any given amount of the substance 1M disintegrate in 40 minutes.
Let the half-life of the substance be taken to be \[{t_{\dfrac{1}{2}}}\].
$t = \dfrac{1}{K}In\dfrac{{[{R^\circ }]}}{{[R]}}$
Where, \[R^\circ \] is the initial concentration = 1 M
R is the remaining concentration \[ = 12.5\% \]
Substituting the values in the above equation, we get
\[40 = \dfrac{1}{K}In\dfrac{{[{R^\circ }]}}{{\dfrac{{12.5}}{{100}}[R]}}\]
$\Rightarrow$ $K = \dfrac{1}{{40}}In\left( {\dfrac{{100}}{{12.5}}} \right)$
$K = \dfrac{1}{{40}}In\,8$
\[{t_{\dfrac{1}{2}}} = ln\dfrac{2}{K}\]
$ = \dfrac{{In\,2}}{{In\dfrac{5}{{40}}}}$
$ = \dfrac{{In\,2}}{{In\,8}} \times 40$
\[ = \dfrac{{40}}{3}\] minutes
= 13.33 minutes
\[{t_{\dfrac{1}{2}}} = 13\] minutes 20 sec
Hence, the half-life of the substance is 13 min 20 sec.
Therefore, the correct answer is option (C).
Note: The time in which half of the original number of nuclei decay is called as the half-life that is “t sec” in this case. Half of the remaining nuclei decay in the following half-life. Therefore, the number of radioactive nuclei decreases from N to \[\dfrac{N}{2}\] in one half-life and $\dfrac{N}{4}$ in the next and to \[\dfrac{N}{8}\] in the next and so on. Radioactive decay is a spontaneous process. It cannot be predicted exactly for any single nucleus and can only be described statistically and probabilistically i.e. can only give averages and probabilities.
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