To determine which isotope could decay to produce [tex]${ }_{82}^{206}Pb$[/tex], you need to think about the characteristics of radioactive decay and the changes in atomic and mass numbers that occur during this process.
1. Radioactive decay: For an isotope to decay into [tex]${ }_{82}^{206}Pb$[/tex], it must start with a higher atomic number (number of protons) and/or mass number (sum of protons and neutrons) because during decay, both of these numbers can decrease. Keep in mind:
- Alpha Decay: The emission of an alpha particle (composed of 2 protons and 2 neutrons) reduces the atomic number by 2 and the mass number by 4.
- Beta Decay: An electron (or positron) emission changes the atomic number by 1 (either increasing or decreasing) without changing the mass number.
- Other decay processes: They typically tweak these numbers in less predictable ways.
Given options:
- 238 (no further context provided, so we will disregard this in lack of specific information).
- [tex]${ }_{87}^{222}Pb$[/tex]
- [tex]${ }_{72}^{178}Hf$[/tex]
- [tex]${ }_{77}^{192}Pb$[/tex]
Let's analyze each option:
### Option: [tex]${ }_{87}^{222}Pb$[/tex]
- Atomic Number: 87
- Mass Number: 222
The element with atomic number 87 is Francium (Fr), not lead (Pb). Assuming this is a hypothetical isotope notation error and continuing:
- 87 > 82: This would have to decay by losing 5 protons.
- 222 > 206: This however, could reasonably decay by losing 16 units in mass (e.g., 4 alpha decays (4 × 4)) or another combination of processes.
- Valid Characteristics: High mass and atomic number sufficient for decay to Pb-206.
### Option: [tex]${ }_{72}^{178}Hf$[/tex]
- Atomic Number: 72 (Hafnium/Hf)
- Mass Number: 178
- 72 < 82: This makes it impossible to decay into Pb-206 as you cannot increase atomic number through nuclear decay.
- Therefore, invalid.
### Option: [tex]${ }_{77}^{192}Pb$[/tex]
- Atomic Number: 77 (Iridium/Ir)
- Mass Number: 192
- 77 < 82: Iridium cannot decay to lead as it requires increasing its atomic number.
- 192 < 206: Again, this isotope doesn't provide enough mass to decay to achieve mass number 206.
- Therefore, invalid.
### Conclusion
The only option where both the atomic number and the mass number start higher and can be reduced through known decay processes to form [tex]${ }_{82}^{206}Pb$[/tex] is option [tex]${ }_{87}^{222}Pb$[/tex].
So, the correct atomic symbol representing an isotope that could undergo radioactive decay to produce [tex]${ }_{82}^{206}Pb$[/tex] is:
[tex]${ }_{87}^{222}Pb$[/tex]
