
Iodine- \[131\] is a radioactive isotope with a half-life of \[8\] days. How many grams of a \[64{\text{ }}g\] sample of iodine - \[131\] will remain at the end of \[24\] days?
Answer
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Hint: Isotopes are those species which have atomic numbers but having different atomic masses. The radioisotopes or the radioactive isotopes are named as radionuclide or radioactive nuclide. These species are those, whose atomic number is same but differs with masses whose nuclei are unstable and vanishes the excess energy by spontaneously emitting radiation in the form of alpha, beta and gamma rays.
Complete step-by-step answer:
Half-life \[(\,{t_{1/2}}\,)\] is the time required for a quantity to reduce into half of its initial value of a compound. The nuclear half-life conveys about how much time passes in the order for half of the atoms present in the initial sample to undergo radioactive decay. Due to this, half-life is important to know.
Half-life says about the time intervals that are expected from radioactive isotopes to be halved.
Suppose, Iodine- \[131\]is having \[8\] days of half-life. Therefore, it denotes that every \[8\] days the sample of Iodine- \[131\] gets halved.
The decay actually follows the half-order kinetics.
\[{t_{12}} = \dfrac{{\ln 2}}{k}\]
The given values are;
Half-life of iodine - \[131\] = \[8\] days
So, we need to find the Amount remaining after \[24\] days.
\[\Rightarrow 8 = \dfrac{{\ln 2}}{k}\]
\[\Rightarrow k = \dfrac{{\ln 2}}{8}\]
Let’s write the first order of kinetics;
\[\Rightarrow \ln \left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \,kt\]
\[\Rightarrow \ln \left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \,\dfrac{{\ln 2}}{8} \cdot 24 = 3\ln 2 = \ln {2^3}\]
\[\Rightarrow \ln \left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \ln 8\]
\[
\Rightarrow \left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \,8 \\
\\
\]
Now, as the concentration is greater the mass will be greater;
\[\Rightarrow C\,\alpha \,M\]
\[\Rightarrow \left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \,\left[ {\dfrac{{{m_0}}}{{{m_t}}}} \right]\, = \,8\]
Total amount of sample = \[64{\text{ }}g\]
\[\Rightarrow {m_0}\, = \,64\,g\]
\[\Rightarrow \,\left[ {\dfrac{{64}}{{{m_t}}}} \right] = \,8\]
\[
\Rightarrow {m_t}\, = \,8\,g \\
\\
\]
So, the sample of iodine - \[131\] will remain \[8\] grams at the end of \[24\] days
Therefore, the answer is \[8\] grams of iodine - \[131\] will remain.
Note: Radioactive decay is the spontaneous process of breakdown of an atomic nucleus which results in the releasing of energy and the matter from the nucleus. The radioisotopes have the unstable nucleus which doesn’t have enough binding energy to hold the nucleus together.
Complete step-by-step answer:
Half-life \[(\,{t_{1/2}}\,)\] is the time required for a quantity to reduce into half of its initial value of a compound. The nuclear half-life conveys about how much time passes in the order for half of the atoms present in the initial sample to undergo radioactive decay. Due to this, half-life is important to know.
Half-life says about the time intervals that are expected from radioactive isotopes to be halved.
Suppose, Iodine- \[131\]is having \[8\] days of half-life. Therefore, it denotes that every \[8\] days the sample of Iodine- \[131\] gets halved.
The decay actually follows the half-order kinetics.
\[{t_{12}} = \dfrac{{\ln 2}}{k}\]
The given values are;
Half-life of iodine - \[131\] = \[8\] days
So, we need to find the Amount remaining after \[24\] days.
\[\Rightarrow 8 = \dfrac{{\ln 2}}{k}\]
\[\Rightarrow k = \dfrac{{\ln 2}}{8}\]
Let’s write the first order of kinetics;
\[\Rightarrow \ln \left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \,kt\]
\[\Rightarrow \ln \left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \,\dfrac{{\ln 2}}{8} \cdot 24 = 3\ln 2 = \ln {2^3}\]
\[\Rightarrow \ln \left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \ln 8\]
\[
\Rightarrow \left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \,8 \\
\\
\]
Now, as the concentration is greater the mass will be greater;
\[\Rightarrow C\,\alpha \,M\]
\[\Rightarrow \left[ {\dfrac{{{A_0}}}{{{A_t}}}} \right] = \,\left[ {\dfrac{{{m_0}}}{{{m_t}}}} \right]\, = \,8\]
Total amount of sample = \[64{\text{ }}g\]
\[\Rightarrow {m_0}\, = \,64\,g\]
\[\Rightarrow \,\left[ {\dfrac{{64}}{{{m_t}}}} \right] = \,8\]
\[
\Rightarrow {m_t}\, = \,8\,g \\
\\
\]
So, the sample of iodine - \[131\] will remain \[8\] grams at the end of \[24\] days
Therefore, the answer is \[8\] grams of iodine - \[131\] will remain.
Note: Radioactive decay is the spontaneous process of breakdown of an atomic nucleus which results in the releasing of energy and the matter from the nucleus. The radioisotopes have the unstable nucleus which doesn’t have enough binding energy to hold the nucleus together.
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