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Chapter 18: Problem 32

When a radon-222 nucleus decays, an alpha particle is emitted. Write the nuclear equation to show what happens when a radon-222 nucleus decays. What is the other product that forms?

Short Answer

Expert verified
The nuclear equation is \(^{222}_{86}\)Rn -> \(^{4}_{2}\)He + \(^{218}_{84}\)Po. The other product formed is a polonium-218 nucleus.

Step by step solution

01

Identify the Initial Particles

The exercise mentions a radon-222 nucleus is decaying. Radon is a chemical element with atomic number 86, represented by Rn in the periodic table. Therefore, the initial particle is \(^{222}_{86}\)Rn.
02

Identify the Emitted Particle

In alpha decay, an alpha particle is emitted. An alpha particle consists of 2 protons and 2 neutrons, which makes it a helium-4 nucleus, represented as \(^{4}_{2}\)He.
03

Determine the Resulting Particle

To determine the resulting particle, subtract the atomic and mass numbers of the emitted alpha particle from the initial radon-222 nucleus. That means the resulting particle has 84 protons and 218 nucleons, which makes it polonium-218 (Po), \(^{218}_{84}\)Po.

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