Consider the following information: i. The layer of dead skin on our bodies is sufficient to protect us from most \(\alpha\) -particle radiation. ii. Plutonium is an \(\alpha\) -particle producer. iii. The chemistry of \(\mathrm{Pu}^{4+}\) is similar to that of \(\mathrm{Fe}^{3+}\). iv. Pu oxidizes readily to \(\mathrm{Pu}^{4+}\) Why is plutonium one of the most toxic substances known?

Short Answer

Expert verified
Plutonium is highly toxic because it readily oxidizes to form \(Pu^{4+}\), which has a similar chemistry to the essential iron ion \(Fe^{3+}\). This similarity allows plutonium to easily enter our cells and tissues, where it emits α-particle radiation, causing significant damage to internal organs and tissues, and leading to serious health problems.

Step by step solution

01

Understand α-particle radiation protection

α-particles are a type of ionizing radiation emitted by some radioactive elements, including plutonium. The layer of dead skin on our bodies offers protection against most α-particle radiation, preventing them from reaching and damaging internal organs and tissues.
02

Recognize plutonium as an α-particle producer

Plutonium is an α-particle producer, meaning that it can emit this type of ionizing radiation. However, as we've already established, our layer of dead skin is usually sufficient to protect us against α-particles.
03

Understand the chemistry of Pu^{4+} and Fe^{3+}

The exercise mentions that the chemistry of Pu^{4+} is similar to that of Fe^{3+}. This is important because the human body requires small amounts of iron (Fe) to function properly. Our cells might not be able to distinguish between the two ions easily, potentially allowing plutonium to enter cells and tissues where it should not be present.
04

Recognize that Pu oxidizes readily to Pu^{4+}

The exercise states that plutonium can readily oxidize to form Pu^{4+}. Since the chemistry of Pu^{4+} is similar to Fe^{3+} and our cells may not be able to differentiate between them, this form of plutonium may be easily absorbed by our body.
05

Connect the facts to explain plutonium's toxicity

Combining the information that (1) plutonium produces α-particle radiation, (2) the chemistry of Pu^{4+} is similar to Fe^{3+}, and (3) plutonium oxidizes readily to form Pu^{4+}, we can conclude that plutonium is highly toxic because it can easily enter our cells and tissues by mimicking the iron our body needs. Once inside the body, the α-particle radiation emitted by plutonium can cause significant damage to internal organs and tissues. This damage leads to serious health problems and makes plutonium one of the most toxic substances known.

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Most popular questions from this chapter

Naturally occurring uranium is composed mostly of \(^{238} \mathrm{U}\) and \(^{235} \mathrm{U},\) with relative abundances of \(99.28 \%\) and \(0.72 \%,\) respectively. The half-life for \(^{238} \mathrm{U}\) is \(4.5 \times 10^{9}\) years, and the half-life for \(^{235} \mathrm{U}\) is \(7.1 \times 10^{8}\) years. Assuming that the earth was formed 4.5 billion years ago, calculate the relative abundances of the \(^{238} \mathrm{U}\) and \(^{235} \mathrm{U}\) isotopes when the earth was formed.

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Using the kinetic molecular theory (see Section 8.6), calculate the root mean square velocity and the average kinetic energy of \(_{1}^{2} \mathrm{H}\) nuclei at a temperature of \(4 \times 10^{7} \mathrm{K}\). (See Exercise 50 for the appropriate mass values.)

The first atomic explosion was detonated in the desert north of Alamogordo, New Mexico, on July \(16,1945 .\) What percentage of the strontium- \(90(t_{1 / 2}=28.9\) years) originally produced . by that explosion still remains as of July \(16,2015 ?\)

In 1994 it was proposed (and eventually accepted) that element 106 be named seaborgium, \(\mathrm{Sg}\), in honor of Glenn T. Seaborg, discoverer of the transuranium elements.a. \(^{263} \mathrm{Sg}\) was produced by the bombardment of \(^{249} \mathrm{Cf}\) with a beam of \(^{18} \mathrm{O}\) nuclei. Complete and balance an equation for this reaction. b. \(^{263}\) Sg decays by \(\alpha\) emission. What is the other product resulting from the \(\alpha\) decay of \(^{263} \mathrm{Sg} ?\)

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