A sample of \displaylines 53 I 131 , as iodide ion was administered to a patient as a carrier consisting of 100.74 mg of stable iodide ion. After 4 days, 66.66% of the initial radioactivity was dete

Question

A sample of \displaylines 53 ​ I 131 , as iodide ion was administered to a patient as a carrier consisting of 100.74 mg of stable iodide ion. After 4 days, 66.66% of the initial radioactivity was detected in the thyroid gland of the patient. What is the approximate mass of the stable iodide ion in mg that has migrated to the thyroid gland? Assume that the rate of migration of stable iodide ion is same as that of radioactive iodide. [ t 1/2 ​ of 131 I = 8.0 days] Given: lo g 2 = 0.301 and antilog of 0.1505 = 2 ​ .

TestHub · Accepted Answer

According to the decay law, the radioactivity of the sample can be determined (this takes into account only the loss of the sample due to decay) as:

$k t = ln (\frac{A _{o}}{A _{t}})$

$\frac{0.693}{8} \times 4 = 2.303 lo g (\frac{A _{o}}{A _{t}})$

$\frac{2.303 l o g 2}{2} = 2.303 lo g \frac{A _{0}}{A _{t}}$

$\frac{0.301}{2} = lo g \frac{A _{0}}{A _{t}}$

$0.1505 = lo g (\frac{A _{o}}{A _{t}})$

$2 = \frac{A _{o}}{A _{t}} = 1.414$

$A_{t} = 0.707 A_{o}$

While the observed activity given takes into account the loss due to decay as well as the loss due to migration of radioactive iodide ion.

So, if activity after 4 days would have been 0.707 A_o, the migration will be 100% but when the activity after 4 days is 0.6666 A_o or 2/3A_o, the migration will be

$\frac{100}{0.707 A _{0}} \times \frac{2}{3} A_{0} = 94.3$ %

As given in the problem, the rate of migration of stable iodide ion is same as that of radioactive iodide, so the mass of stable iodide ion that has migrated to thyroid gland will be:

$100.74 m g \times \frac{94.3}{100} \approx 94.9 m g \approx 95 m g$

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