Q. Half-life of radioactive substance is $3.20\, hr$. What is the time taken for $75 \%$ of substance to be used?

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A

$6.4\, hr$

B

$12 \, hr$

C

$4.18$ day

D

$1.2$ day

Solution:

$N_{0} \underset{1}{\rightarrow} \frac{N_{0}}{2} \underset{2}{\rightarrow} \frac{N_{o}}{4}$
Remaining substance after two half-lives is $\frac{N_{0}}{4}$
(or) the substance used during this time
$=N_{0}-\frac{N_{0}}{4}=3 \frac{N_{0}}{4}=75 \% N_{0}$
Therefore, time taken is two half-lives
$\Rightarrow t=2 \times 3.2=6.4\, hr$.

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