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Question: Half-life of a radioactive substance is\(3.20\,{\text{h}}\). What is the time taken for a \(75\% \)o...

Half-life of a radioactive substance is3.20 h3.20\,{\text{h}}. What is the time taken for a 75%75\% of substance to be used?
A. 1.21.2days
B. 4.184.18days
C. 6.38 h6.38\,{\text{h}}
D. 12 h12\,{\text{h}}

Explanation

Solution

Half-life related to a radioactive material is the time period needed for the decay of one-half of the atomic nuclei in a radioactive sample.
The formula to calculate the time period is,
RRo=(12)t/T1/2\dfrac{R}{{{R_o}}} = {\left( {\dfrac{1}{2}} \right)^{t/{T_{1/2}}}}
Where Ro{{\text{R}}_{\text{o}}}refers to the starting quantity of the isotope that will decay, RRis the remaining quantity, T1/2{T_{1/2}}represents half-life of the decaying quantity and ttis the time taken.

Complete step by step solution:
Half-life as the name denotes means half of a particular substance to react chemically.
It is a common term in nuclear physics and the survival of atoms in radioactive decay depends on the half-life. Half-life can encourage the classification of any form of decay whether rapidly changing or not.

Given,
Half-life of the radioactive substance,
T1/2{T_{{\text{1/2}}}} =3.20 h = 3.20\,{\text{h}}
Percentage of the substance used =75 = 75

Therefore, percentage of substance left

= \dfrac{R}{{{R_o}}} \\\ = 1 - \dfrac{{75}}{{100}} \\\ = 25\% \\\ \end{gathered} $$ The time taken for $75\% $of the substance to be used, $t = ?$ According to the formula, $\dfrac{R}{{{R_o}}}{\text{ = }}{\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)^{{\text{t}}/{T_{{\text{1/2}}}}}}$ $\dfrac{{25}}{{100}} = {\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)^{t/{T_{{\text{1/2}}}}}}$ ${\left( {\dfrac{1}{2}} \right)^2} = {\left( {\dfrac{{\text{1}}}{{\text{2}}}} \right)^{{\text{t}}/{T_{{\text{1/2}}}}}}$ $\dfrac{t}{{{T_{{\text{1/2}}}}}} = {\text{2}}$ Therefore, $t = {\text{2 \times }}{T_{{\text{1/2}}}}$ $ = 2 \times 3.20$ $ = 6.40\, \approx {\text{6}}{\text{.38}}\,{\text{h}}$ **Hence, time taken, ${\text{t}}$ for $75\% $ of the substance to be used is${\text{6}}{\text{.38}}\,{\text{h}}$.** **So, option C is correct.** **Note:** There may be many half-lives depending on the reaction. A radio-active substance may have first half-life, second half-life etc and we have to change the calculations based on that. A specific radioactive material has a steady half-life.