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Chapter 2: Problem 38

Iodine-131 is a radioactive isotope that has important medical uses. Small doses of iodine-131 are used for treating hyperthyroidism (overactive thyroid) and larger doses are used for treating thyroid cancer. Iodine-131 is administered to patients in the form of sodium iodide capsules that contain \(^{131} \mathrm{I}^{-}\) ions. Determine the number of neutrons, protons, and electrons in a single \(^{131} \mathrm{I}^{-}\) ion.

Short Answer

Expert verified
In a single iodine-131 ion, there are 53 protons, 78 neutrons, and 54 electrons.

Step by step solution

01

Determining the number of Protons

The atom in question is Iodine (\(\mathrm{I}\)). The atomic number of iodine, which is given by its place in the periodic table, is 53. This represents the number of protons. Thus, the number of protons in iodine-131 is 53.
02

Determining the number of Neutrons

To find the number of neutrons in a nucleus, the atomic number (No. of protons) is subtracted from the mass number. In this case, Iodine-131, the atomic mass is 131 (given in the notation). So, the number of neutrons = atomic mass – atomic number = \(131 - 53 = 78\). Therefore, there are 78 neutrons in a single iodine-131 isotope.
03

Determining the number of Electrons

In an uncharged, neutral atom, the number of electrons is equal to the number of protons. However, the iodine in the form given in the problem (\(^{131} \mathrm{I}^{-}\)) is an ion with a negative charge. This means it has gained one extra electron. Therefore, the number of electrons = atomic number + 1 (due to the negative charge) = \(53 + 1 = 54\). Thus, there are 54 electrons in a single iodine-131 ion.

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Most popular questions from this chapter

Iodine has many radioactive isotopes. Iodine-123 is a radioactive isotope used for obtaining images of the thyroid gland. Iodine-123 is administered to patients in the form of sodium iodide capsules that contain \(123 \mathrm{I}^{-}\) ions. Determine the number of neutrons, protons, and electrons in a single \(^{123} \mathrm{I}^{-}\) ion.

There are three naturally occurring isotopes of magnesium. Their masses and percent natural abundances are \(23.985042 \mathrm{u}, 78.99 \% ; 24.985837 \mathrm{u}, 10.00 \% ;\) and \(25.982593 \mathrm{u}, 11.01 \% .\) Calculate the weighted-average atomic mass of magnesium.

The following data on isotopic masses are from a chemical handbook. What is the ratio of each of these masses to that of\(^{12}_{6} \mathrm{c}\) ?\( \)^{17}_{35} \mathrm{CL}\( , 34.96885 \mathrm{u} ;\) (b) \(_{12}^{26} \mathrm{Mg}\) 25.98259 u ;\(^{86}_{222} \mathrm{Rn}\) , 222.0175 u.

A 0.406 g sample of magnesium reacts with oxygen, producing \(0.674 \mathrm{g}\) of magnesium oxide as the only product. What mass of oxygen was consumed in the reaction?

The German chemist Fritz Haber proposed paying off the reparations imposed against Germany after World War I by extracting gold from seawater. Given that (1) the amount of the reparations was \(\$ 28.8\) billion dollars, (2) the value of gold at the time was about \(\$ 21.25\) per troy ounce ( \(1 \text { troy ounce }=31.103 \mathrm{g}),\) and (3) gold occurs in seawater to the extent of \(4.67 \times 10^{17}\) atoms per ton of seawater \((1 \text { ton }=2000\) lb), how many cubic kilometers of seawater would have had to be processed to obtain the required amount of gold? Assume that the density of seawater is \(1.03 \mathrm{g} / \mathrm{cm}^{3}\) (Haber's scheme proved to be commercially infeasible, and the reparations were never fully paid.)
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