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Chapter 2: Problem 28

For the ion \(^{228} \mathrm{Ra}^{2+}\) with a mass of 228.030 u, determine (a) the numbers of protons, neutrons, and electrons in the ion; (b) the ratio of the mass of this ion to that of an atom of \(^{16} \mathrm{O}\) (refer to page 47 ).

Short Answer

Expert verified
The ion \(^{228} \mathrm{Ra}^{2+}\) has 88 protons, 140 neutrons, and 86 electrons. The ratio of its mass to that of an Oxygen atom is approximately 14.25.

Step by step solution

01

Numbers of protons, neutrons, and electrons

In \(^{228} \mathrm{Ra}^{2+}\), 'Ra' stands for Radium and its atomic number is 88. This means it has 88 protons. Since it's a \(^{2+}\) ion, it has lost two electrons, so it has 88 - 2 = 86 electrons. The atomic mass (A) is 228, so the number of neutrons can be calculated by A - atomic number (Z), which is 228 - 88 = 140 neutrons.
02

Mass ratio calculation

The ratio of the mass of the ion to that of an atom of \(^{16} \mathrm{O}\) is simply the mass of \(^{228} \mathrm{Ra}^{2+}\) divided by the mass of \(^{16} \mathrm{O}\). Given that the atomic mass unit (u) of Oxygen is approximately 16 u, the ratio would be 228.030 / 16 = 14.251875.

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