
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
547.8k+ views
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.
Recently Updated Pages
A man running at a speed 5 ms is viewed in the side class 12 physics CBSE

State and explain Hardy Weinbergs Principle class 12 biology CBSE

Which of the following statements is wrong a Amnion class 12 biology CBSE

Two Planoconcave lenses 1 and 2 of glass of refractive class 12 physics CBSE

The compound 2 methyl 2 butene on reaction with NaIO4 class 12 chemistry CBSE

Bacterial cell wall is made up of A Cellulose B Hemicellulose class 12 biology CBSE

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Give 10 examples of unisexual and bisexual flowers

State the principle of an ac generator and explain class 12 physics CBSE

