Question

complete the nuclear equation and determine the type of decay that is occurring in this reaction.

Explanation:

Step1: Determine the mass number and atomic number of the missing particle

The mass number of \(^{226}_{88}\text{Ra}\) is \(226\), and the mass number of \(^{222}_{86}\text{Rn}\) is \(222\). So the mass number of the missing particle is \(226 - 222=4\).
The atomic number of \(^{226}_{88}\text{Ra}\) is \(88\), and the atomic number of \(^{222}_{86}\text{Rn}\) is \(86\). So the atomic number of the missing particle is \(88 - 86 = 2\).

Step2: Identify the particle

A particle with mass number \(4\) and atomic number \(2\) is an alpha particle (\(^{4}_{2}\text{He}\)).

Step3: Determine the type of decay

Since an alpha particle (\(^{4}_{2}\text{He}\)) is emitted, this is alpha decay.

Answer:

The completed nuclear equation is \(^{226}_{88}\text{Ra}\to^{222}_{86}\text{Rn}+^{4}_{2}\text{He}\), and the type of decay is alpha decay.