In the given nuclear reaction:
[tex]$
{}_{86}^{222}Rn \rightarrow {}_{84}^{218}Po + ?
$[/tex]
We start by determining the changes in both the atomic number (number of protons) and the mass number (total number of protons and neutrons).
1. Identify the properties of the original isotope (Rn-222):
- Atomic number of Rn (Radon): 86 (i.e., 86 protons)
- Mass number of Rn (Radon): 222 (i.e., the total number of protons and neutrons)
2. Identify the properties of the resultant isotope (Po-218):
- Atomic number of Po (Polonium): 84 (i.e., 84 protons)
- Mass number of Po (Polonium): 218 (i.e., the total number of protons and neutrons)
3. Calculate the difference in atomic numbers (number of protons):
- Initial number of protons in Rn: 86
- Final number of protons in Po: 84
- Difference in number of protons: \(86 - 84 = 2\)
4. Calculate the difference in mass numbers (total number of protons and neutrons):
- Initial mass number of Rn: 222
- Final mass number of Po: 218
- Difference in mass numbers: \(222 - 218 = 4\)
5. Interpret the differences:
- The difference in atomic number (2 protons) indicates that 2 protons are released.
- The difference in mass number (4 units) indicates that the total number of particles (protons and neutrons) released is 4. Given that we already have 2 protons accounted for, the remaining 2 units must be neutrons.
An alpha particle consists of 2 protons and 2 neutrons. The changes observed (release of 2 protons and 2 neutrons) match the characteristics of an alpha particle.
Therefore, the particle released in the reaction is:
one alpha particle
