Half-life of a radioactive substance is$3.20\,{\text{h}}$. What is the time taken for a $75\% $of substance to be used?
A. $1.2$days
B. $4.18$days
C. $6.38\,{\text{h}}$
D. $12\,{\text{h}}$
Answer
615.9k+ views
Hint: 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,
$\dfrac{R}{{{R_o}}} = {\left( {\dfrac{1}{2}} \right)^{t/{T_{1/2}}}}$
Where ${{\text{R}}_{\text{o}}}$refers to the starting quantity of the isotope that will decay, $R$is the remaining quantity, ${T_{1/2}}$represents half-life of the decaying quantity and $t$is 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,
${T_{{\text{1/2}}}}$$ = 3.20\,{\text{h}}$
Percentage of the substance used $ = 75$
Therefore, percentage of substance left
\[\begin{gathered}
= \dfrac{R}{{{R_o}}} \\
= 1 - \dfrac{{75}}{{100}} \\
= 25\% \\
\end{gathered} \]
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.
The formula to calculate the time period is,
$\dfrac{R}{{{R_o}}} = {\left( {\dfrac{1}{2}} \right)^{t/{T_{1/2}}}}$
Where ${{\text{R}}_{\text{o}}}$refers to the starting quantity of the isotope that will decay, $R$is the remaining quantity, ${T_{1/2}}$represents half-life of the decaying quantity and $t$is 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,
${T_{{\text{1/2}}}}$$ = 3.20\,{\text{h}}$
Percentage of the substance used $ = 75$
Therefore, percentage of substance left
\[\begin{gathered}
= \dfrac{R}{{{R_o}}} \\
= 1 - \dfrac{{75}}{{100}} \\
= 25\% \\
\end{gathered} \]
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.
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