Problem 65 - Chemistry: The Central Science

Question

Radon-222 decays to a stable nucleus by a series of three alpha emissions and two beta emissions. What is the stable nucleus that is formed?

Step-by-Step Solution

Answer

The stable nucleus formed after three alpha emissions and two beta emissions from Radon-222 is Lead-210, represented by \(^{210}_{82}\)Pb.

Step 1: Identify the initial nucleus

Initially, we have a Radon-222 nucleus. Radon is a chemical element with the atomic number 86 and has a mass number of 222.

Step 2: Alpha emissions

An alpha emission involves the ejection of an alpha particle from the nucleus, consisting of 2 protons and 2 neutrons. After each alpha emission, the atomic number decreases by 2, and the mass number decreases by 4. Since there are three alpha emissions, perform this calculation three times: Initial: \(^{222}_{86}\)Rn 1st alpha emission: \(^{218}_{84}\)Po (Polonium) 2nd alpha emission: \(^{214}_{82}\)Pb (Lead) 3rd alpha emission: \(^{210}_{80}\)Hg (Mercury)

Step 3: Beta emissions

A beta emission involves the conversion of a neutron into a proton, resulting in an increase of the atomic number by 1, while the mass number remains the same. Since there are two beta emissions, perform this calculation twice: After 1st beta emission: \(^{210}_{81}\)Tl (Thallium) After 2nd beta emission: \(^{210}_{82}\)Pb (Lead)

Step 4: Identify the stable nucleus

After all three alpha emissions and two beta emissions, the final nucleus formed is a stable Lead-210 nucleus, represented by: \(^{210}_{82}\)Pb.