Let's analyze the given nuclear reaction:
[tex]\[ _{86}^{222}Rn \rightarrow _{84}^{218}Po + ? \][/tex]
Here, Radon-222 is decaying into Polonium-218. We need to determine the nature of the particle released in this process.
1. Atomic Number Change:
- Initial atomic number of Radon (Rn): 86
- Final atomic number of Polonium (Po): 84
- Change in atomic number: [tex]\( 86 - 84 = 2 \)[/tex]
2. Mass Number Change:
- Initial mass number of Radon (Rn): 222
- Final mass number of Polonium (Po): 218
- Change in mass number: [tex]\( 222 - 218 = 4 \)[/tex]
The change in atomic number indicates the loss of 2 protons, and the change in mass number indicates the loss of 4 nucleons (protons + neutrons).
An alpha particle ([tex]\( \alpha \)[/tex] particle) consists of:
- 2 protons
- 2 neutrons
Therefore, one alpha particle has:
- Atomic number = 2
- Mass number = 4
Given the changes in atomic and mass numbers in this reaction (2 protons and 4 nucleons), this corresponds exactly to an alpha particle.
Thus, the particle released in this reaction is:
One alpha particle.
In conclusion, the correct choice is:
- One alpha particle
