
How long will it take for $\dfrac{3}{4}$ of a sample of 131 iodine that has a half-life of 8.1 days?
Answer
538.5k+ views
Hint: In radioactive elements, half-life is the term used for the time at which the atomic nuclei have been decayed into half of its total mass. Also it is the half course that a radioactive nucleus completes in its decay.
Formula used:
Exponential decay or nuclear half life calculation-
$A(t)={{A}_{0}}.{{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{{{t}_{{}^{1}/{}_{2}}}}}}$
Complete answer:
As the radioactive decay of elements occurs, when they reach a stage where their mass, activity and life becomes half, then half-life is calculated that tells us the life of that atom or when the complete decay will occur.
Now, to calculate the time or life for $\dfrac{3}{4}$ of a sample of 131 iodine, we will put the given quantities in the exponential decay equation as:
$A(t)={{A}_{0}}.{{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{{{t}_{{}^{1}/{}_{2}}}}}}$ where,
A(t)= amount of substance left after (t) years.
${{A}_{0}}$ = the quantity of substance initially before the decay.
t= life of the substance
${{t}_{{}^{1}/{}_{2}}}$ = half-life of the substance
We know that $\dfrac{3}{4}$ of the sample undergoes decay, so, the sample that will be left after the decay will be $\dfrac{1}{4}$.
So, the amount left at this time will be $A(t)={{A}_{0}}.\left( \dfrac{1}{4} \right)$. Putting this value in the exponential decay or half life equation, we get,
${{A}_{0}}\left( \dfrac{1}{4} \right)={{A}_{0}}.{{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{{{t}_{{}^{1}/{}_{2}}}}}}$
So, $\left( \dfrac{1}{4} \right)={{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{{{t}_{{}^{1}/{}_{2}}}}}}$
This shows that $\dfrac{t}{{{t}_{{}^{1}/{}_{2}}}}=2$, as the square root of 2 is 4.
So, $t=2\times {{t}_{{}^{1}/{}_{2}}}$
Given the half life of substance, ${{t}_{{}^{1}/{}_{2}}}=8.1$days
Time taken will be, $t=2\times 8.1\,days$
$t=16.2\,days$
Hence, $\dfrac{3}{4}$ of the sample of 131 iodine will take 16.2 days for its decay.
Note:
Nuclear half-life of the substance can also be calculated from the rate constant of the reaction of the radioactive decay as:
\[\ln \dfrac{{{A}_{t}}}{{{A}_{0}}}=-kt\] where, ${{A}_{t}}$ is substance at time (t), ${{A}_{0}}$is substance initially before decay, t is the time taken, and k is the rate constant of the reaction.
Formula used:
Exponential decay or nuclear half life calculation-
$A(t)={{A}_{0}}.{{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{{{t}_{{}^{1}/{}_{2}}}}}}$
Complete answer:
As the radioactive decay of elements occurs, when they reach a stage where their mass, activity and life becomes half, then half-life is calculated that tells us the life of that atom or when the complete decay will occur.
Now, to calculate the time or life for $\dfrac{3}{4}$ of a sample of 131 iodine, we will put the given quantities in the exponential decay equation as:
$A(t)={{A}_{0}}.{{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{{{t}_{{}^{1}/{}_{2}}}}}}$ where,
A(t)= amount of substance left after (t) years.
${{A}_{0}}$ = the quantity of substance initially before the decay.
t= life of the substance
${{t}_{{}^{1}/{}_{2}}}$ = half-life of the substance
We know that $\dfrac{3}{4}$ of the sample undergoes decay, so, the sample that will be left after the decay will be $\dfrac{1}{4}$.
So, the amount left at this time will be $A(t)={{A}_{0}}.\left( \dfrac{1}{4} \right)$. Putting this value in the exponential decay or half life equation, we get,
${{A}_{0}}\left( \dfrac{1}{4} \right)={{A}_{0}}.{{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{{{t}_{{}^{1}/{}_{2}}}}}}$
So, $\left( \dfrac{1}{4} \right)={{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{{{t}_{{}^{1}/{}_{2}}}}}}$
This shows that $\dfrac{t}{{{t}_{{}^{1}/{}_{2}}}}=2$, as the square root of 2 is 4.
So, $t=2\times {{t}_{{}^{1}/{}_{2}}}$
Given the half life of substance, ${{t}_{{}^{1}/{}_{2}}}=8.1$days
Time taken will be, $t=2\times 8.1\,days$
$t=16.2\,days$
Hence, $\dfrac{3}{4}$ of the sample of 131 iodine will take 16.2 days for its decay.
Note:
Nuclear half-life of the substance can also be calculated from the rate constant of the reaction of the radioactive decay as:
\[\ln \dfrac{{{A}_{t}}}{{{A}_{0}}}=-kt\] where, ${{A}_{t}}$ is substance at time (t), ${{A}_{0}}$is substance initially before decay, t is the time taken, and k is the rate constant of the reaction.
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