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Chapter 21: Problem 62

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
Taking potassium iodide in the event of a nuclear accident may protect against thyroid cancer by filling the thyroid's iodine receptors, preventing any subsequently encountered radioactive iodine from being absorbed.

Step by step solution

01

Understanding the role of iodine in the human body

The thyroid gland in the human body absorbs iodine from the food we eat. This iodine is used to make thyroid hormones that regulate the body's metabolism and other important processes.
02

Realizing the danger of radioactive iodine

In the event of a nuclear accident, radioactive iodine could be released. If this radioactive iodine is absorbed by the bodies of people in close proximity, it could lead to radiation sickness or thyroid cancer.
03

Understanding the role of potassium iodide

Potassium iodide (KI) can be used to protect the body from radioactive iodine. The non-radioactive iodine in KI fills up iodine receptors in the thyroid, ensuring that if any radioactive iodine is subsequently encountered, it cannot be absorbed.
04

Grasping the timing

The important caveat here is that KI needs to be taken before or immediately after exposure to radiation for it to have a protective effect - it is not a cure if radioactive iodine has already been absorbed.

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Most popular questions from this chapter

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?

These 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) \({ }_{92}^{235} \mathrm{U}+{ }_{0}^{1} \mathrm{n} \longrightarrow{ }_{62}^{160} \mathrm{Sm}+{ }_{30}^{72} \mathrm{Zn}+4 \mathrm{X}\)

How do nuclear reactions differ from ordinary chemical reactions?

The radioactive isotope \({ }^{238} \mathrm{Pu},\) used in pacemakers, decays by emitting an alpha particle with a half-life of 86 yr. (a) Write an equation for the decay process. (b) The energy of the emitted alpha particle is \(9.0 \times 10^{-13} \mathrm{~J}\), which is the energy per decay. Assume that all the alpha particle energy is used to run the pacemaker, calculate the power output at \(t=0\) and \(t=10 \mathrm{yr}\). Initially \(1.0 \mathrm{mg}\) of \({ }^{238} \mathrm{Pu}\) was present in the pacemaker (Hint: After \(10 \mathrm{yr}\), the activity of the isotope decreases by 8.0 percent. Power is measured in watts or \(\mathrm{J} / \mathrm{s}\).).

A freshly isolated sample of \(90 \mathrm{Y}\) was found to have an activity of \(9.8 \times 10^{5}\) disintegrations per minute at 1: 00 P.M. on December \(3,2000 .\) At 2: 15 P.M. on December \(17,2000,\) its activity was redetermined and found to be \(2.6 \times 10^{4}\) disintegrations per minute. Calculate the half-life of \(90 \mathrm{Y}\).
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