
The half-life of $^{58}Co$ is 72 days. How many days does it take for $3/4$ of the initial mass to disintegrate?
A. 144 days
B. 14 days
C. 140 days
D. 130 days
Answer
464.7k+ views
Hint: We can see that two half lives have been covered $\left( {\dfrac{3}{4} = \dfrac{1}{2} + \dfrac{1}{4}} \right)$ . We know the time of one half life, so the time taken can be calculated.
Formula used:
The formula for the half life of a substance is
$N = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{{t_{1/2}}}}}}$
where,
$N = $ the initial amount of substance
${N_0} = $ amount of substance still remaining/ amount not yet decayed after time t
${t_{1/2}} = $ half-life of the substance
Complete step by step answer:
here we are given that 3/4 of the substance disintegrates, therefore, the remaining amount of substance is 1/4 th of original quantity. Let the original quantity be N, then the amount left is 1/4 N.
Also, the half life of given isotope of Cobalt is 72 days i.e ${t_{1/2}} = {72\,days}$
Using the formula for half-life,
\[
\dfrac{1}{4}N = N{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{72}}}} \\
\Rightarrow {\left( {\dfrac{1}{2}} \right)^2} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{72}}}} \\
\Rightarrow 2 = \dfrac{t}{{72}} \\
\therefore t = {144\,days} \\
\]
Therefore, the answer is option A.
Additional information:
The half-life of a radioactive substance is a characteristic constant. It measures the time it takes for a given amount of the substance to become reduced by half as a consequence of decay, and therefore, the emission of radiation.Archeologists and geologists use half-life to date the age of organic objects in a process known as carbon dating. During beta decay, carbon 14 becomes nitrogen 14. At the time of death organisms stop producing carbon 14. Since half life is a constant, the ratio of carbon 14 to nitrogen 14 provides a measurement of the age of a sample.
Note:We can also simply see that 3/4 of the original amount will disintegrate after 2 half lives because one half life means 1/2 of the substance and 1/4 means the second half life. And two half lives mean two times the half-life period. 72x2=144days
Formula used:
The formula for the half life of a substance is
$N = {N_0}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{{t_{1/2}}}}}}$
where,
$N = $ the initial amount of substance
${N_0} = $ amount of substance still remaining/ amount not yet decayed after time t
${t_{1/2}} = $ half-life of the substance
Complete step by step answer:
here we are given that 3/4 of the substance disintegrates, therefore, the remaining amount of substance is 1/4 th of original quantity. Let the original quantity be N, then the amount left is 1/4 N.
Also, the half life of given isotope of Cobalt is 72 days i.e ${t_{1/2}} = {72\,days}$
Using the formula for half-life,
\[
\dfrac{1}{4}N = N{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{72}}}} \\
\Rightarrow {\left( {\dfrac{1}{2}} \right)^2} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{72}}}} \\
\Rightarrow 2 = \dfrac{t}{{72}} \\
\therefore t = {144\,days} \\
\]
Therefore, the answer is option A.
Additional information:
The half-life of a radioactive substance is a characteristic constant. It measures the time it takes for a given amount of the substance to become reduced by half as a consequence of decay, and therefore, the emission of radiation.Archeologists and geologists use half-life to date the age of organic objects in a process known as carbon dating. During beta decay, carbon 14 becomes nitrogen 14. At the time of death organisms stop producing carbon 14. Since half life is a constant, the ratio of carbon 14 to nitrogen 14 provides a measurement of the age of a sample.
Note:We can also simply see that 3/4 of the original amount will disintegrate after 2 half lives because one half life means 1/2 of the substance and 1/4 means the second half life. And two half lives mean two times the half-life period. 72x2=144days
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