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# A human body requires the $0.01M$ activity of radioactive substance after $24h$. Half-life of radioactive substance is $6h$,Then injection of maximum activity of radioactive substance that can be injected is:A.$0.08$B.$0.04$C.$0.16$D.$0.32$

Last updated date: 23rd Feb 2024
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Hint: We first need to discover the quantity of half-life that will sit back from rotting. At that point utilizing the formula, we will get the part of substance remaining and by taking it away from one, we will get the necessary solution.
Formula used:
Division of a substance staying after $n$ half-lives $= \dfrac{1}{{{2^n}}}$

Half-life ( ${t_{1 / 2}}$) is the time needed for an amount to decrease to half of its underlying worth. The term is regularly utilized in atomic material science to portray how rapidly flimsy particles go through radioactive rot or how long stable molecules endure. The term is additionally utilized all the more by and large to describe any kind of remarkable or non-outstanding rot. For instance, the clinical sciences allude to the organic half-life of medications and different synthetic compounds in the human body.
We have been given the half-life of the radioactive substance, ${t_{\dfrac{1}{2}}} = 6$ minutes.
Furthermore, we need to discover the level of substance rotted following $24$ minutes.
Thusly, number of half-lives passed, $n = \dfrac{{24}}{6} = 4$
Remaining activity $= 0.01M$ after $24h$
$Remaining{\text{ }}activity = {\text{ }}Initial{\text{ }}activity{\text{ }} \times {\left( {\dfrac{1}{2}} \right)^n}$
Thus, $0.01 = {\text{ }}Initial{\text{ }}activity{\text{ }} \times {\left( {\dfrac{1}{2}} \right)^4}$
Initial activity $= 0.01 \times 16 = 0.16{\text{ }}M$
Thus, the right choice is ($C$).