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Chapter 19: Problem 67

After the Chernobyl accident, people living close to the nuclear reactor site were urged to take large amounts of potassium iodide as a safety precaution. What is the chemical basis for this action?

Short Answer

Expert verified
Potassium iodide provides a source of stable iodine that saturates the thyroid gland, preventing it from absorbing harmful radioactive iodine that may be released during a nuclear accident. Therefore, potassium iodide is taken as a safety precaution to protect the thyroid gland from damage by radioactive iodine.

Step by step solution

01

Understanding the Role of Iodine in the Body

Iodine is an essential element that the body cannot produce, but it's very important for the thyroid gland. It produces thyroid hormones, which maintain body metabolism, growth, and development. Normally, iodine, which the body obtains from certain foods, is absorbed by the thyroid gland for the production of these hormones.
02

Comprehension of Radioactive Iodine Exposure

In the event of a nuclear accident, one type of radioactive material that can be released into the atmosphere is radioactive iodine (iodine-131). If inhaled or swallowed, it can be absorbed by the thyroid gland, damaging its cells and causing diseases such as cancer.
03

The Protective Role of Potassium Iodide

Consuming non-radioactive potassium iodide prior or immediately after exposure to radiation can protect the thyroid from damage. The gland quickly absorbs it, saturating itself with stable, non-radioactive iodine. This millieu effectively blocks the uptake of radioactive iodine-131, which is then excreted from the body.

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Most popular questions from this chapter

The following equations are for nuclear reactions that are known to occur in the explosion of an atomic bomb. Identify X. (a) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{56}^{140} \mathrm{Ba}+3{ }_{0}^{1} \mathrm{n}+\mathrm{X}\) (b) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{55}^{144} \mathrm{Cs}+{ }_{37}^{90} \mathrm{Rb}+2 \mathrm{X}\) (c) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{35}^{87} \mathrm{Br}+3{ }_{0}^{1} \mathrm{n}+\mathrm{X}\) (d) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{62}^{160} \mathrm{Sm}+{ }_{30}^{72} \mathrm{Zn}+4 \mathrm{X}\).

Consider the decay series $$\mathrm{A} \longrightarrow \mathrm{B} \longrightarrow \mathrm{C} \longrightarrow \mathrm{D}$$ where \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are radioactive isotopes with halflives of \(4.50 \mathrm{~s}, 15.0\) days, and \(1.00 \mathrm{~s},\) respectively, and \(\mathrm{D}\) is nonradioactive. Starting with 1.00 mole of A, and none of \(\mathrm{B}, \mathrm{C},\) or \(\mathrm{D},\) calculate the number of moles of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) left after 30 days.

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