Answer
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Hint: Activity in a radioactive reaction is the decay of unstable nuclei per second or it can be referred to as the rate of decay for that radioactive reaction.
-As we know the radioactive decay is a type of first order reaction this question can be solved by using the same formula and method used to calculate the activity for simple non-radioactive chemical reactions of first order.
Formula Used:
$\lambda = \dfrac{{2.303}}{t}\log \dfrac{{{N_0}}}{N}$
Here $\lambda $ is the rate of decay
${N_0}$is initial value of unstable nuclei
$N$ is the value of unstable nuclei after time $t$
Also, for a first order reaction
$\lambda = \dfrac{{0.693}}{{{t_{1/2}}}}$
Here ${t_{1/2}}$ is the time for the unstable nuclei to decay by $50\% $ of their initial value
Complete step by step solution:
Now for the given question ${t_{1/2}}$ for iodide ion is given as eight days using that in the following formula we can calculate the decay for iodine ion as follow
$\lambda = \dfrac{{0.693}}{{{t_{1/2}}}}$
$\lambda = \dfrac{{0.693}}{8}$ (i)
Also
$\lambda = \dfrac{{2.303}}{t}\log \dfrac{{{N_0}}}{N}$ (ii)
Here $t$ is given as $4$ days
Equating both the equations (i) and (ii)
We get
$\dfrac{{0.693}}{8} = \dfrac{{2.303}}{t}\log \dfrac{{{N_0}}}{N}$
$\dfrac{N}{{{N_0}_{}}} = 0.707$
This means that $70.7\% $ of the initial number of unstable nuclei are present.
While $66.67\% $ of the initial radioactivity was detected in the thyroid gland of the patient
Hence Weight of iodine migrated to the thyroid gland is $\dfrac{{66.67}}{{70.7}} \times 0.1\,mg$
This comes out as $0.0958\,mg$
Hence the option ‘A’ is the correct solution for the given question.
Note:
$^{131}{I_{53}}$ is an isotope of iodine that emits radiation. When a small dose of $I - 131$ is swallowed, it is absorbed into the bloodstream in the gastrointestinal (GI) tract and concentrated from the blood by the thyroid gland, where it begins destroying the gland's cells
Thyroid is a disease caused due to deficiency of iodine in the human body.
-As we know the radioactive decay is a type of first order reaction this question can be solved by using the same formula and method used to calculate the activity for simple non-radioactive chemical reactions of first order.
Formula Used:
$\lambda = \dfrac{{2.303}}{t}\log \dfrac{{{N_0}}}{N}$
Here $\lambda $ is the rate of decay
${N_0}$is initial value of unstable nuclei
$N$ is the value of unstable nuclei after time $t$
Also, for a first order reaction
$\lambda = \dfrac{{0.693}}{{{t_{1/2}}}}$
Here ${t_{1/2}}$ is the time for the unstable nuclei to decay by $50\% $ of their initial value
Complete step by step solution:
Now for the given question ${t_{1/2}}$ for iodide ion is given as eight days using that in the following formula we can calculate the decay for iodine ion as follow
$\lambda = \dfrac{{0.693}}{{{t_{1/2}}}}$
$\lambda = \dfrac{{0.693}}{8}$ (i)
Also
$\lambda = \dfrac{{2.303}}{t}\log \dfrac{{{N_0}}}{N}$ (ii)
Here $t$ is given as $4$ days
Equating both the equations (i) and (ii)
We get
$\dfrac{{0.693}}{8} = \dfrac{{2.303}}{t}\log \dfrac{{{N_0}}}{N}$
$\dfrac{N}{{{N_0}_{}}} = 0.707$
This means that $70.7\% $ of the initial number of unstable nuclei are present.
While $66.67\% $ of the initial radioactivity was detected in the thyroid gland of the patient
Hence Weight of iodine migrated to the thyroid gland is $\dfrac{{66.67}}{{70.7}} \times 0.1\,mg$
This comes out as $0.0958\,mg$
Hence the option ‘A’ is the correct solution for the given question.
Note:
$^{131}{I_{53}}$ is an isotope of iodine that emits radiation. When a small dose of $I - 131$ is swallowed, it is absorbed into the bloodstream in the gastrointestinal (GI) tract and concentrated from the blood by the thyroid gland, where it begins destroying the gland's cells
Thyroid is a disease caused due to deficiency of iodine in the human body.
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